ICSE Class 10 Mathematics Chapter 8 Selina Concise Solutions

1. Find the remainder when ${x^2} - 8x + 4$ is divided by $2x + 1$.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

$2x + 1 = 0 \Rightarrow x =  - \frac{1}{2}$

Required remainder $= $ Value of given polynomial ${x^2} - 8x + 4$ at $x =  - \frac{1}{2}$

${\text{Remainder }} = {\left( { - \frac{1}{2}} \right)^2} - 8\left( { - \frac{1}{2}} \right) + 4$

$= \frac{1}{4} + 4 + 4$

$= 8\frac{1}{4}$

2. Find the value of ' $a$ ' if the division of $a{x^3} + 9{x^2} + 4x - 10$ by $x + 3$ leaves a remainder of ${\mathbf{5}}$.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

$x + 3 = 0 \Rightarrow x =  - 3$

Given, the remainder is $5$.

Thus, the value of $a{x^3} + 9{x^2} + 4x - 10$ at $x =  - 3$ is $5$

$a{( - 3)^3} + 9{( - 3)^2} + 4( - 3) - 10 = 5$

$ \Rightarrow  - 27a + 81 - 12 - 10 = 5$

Then, $a = 2$

3. When the polynomial $2{x^3} - k{x^2} + (5k - 3)x - 8$ is divided by $x - 2$, the remainder is $14$ . Find the value of ' $k$ '.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

$x - 2 = 0 \Rightarrow x = 2$

Given, the remainder is $14$.

Therefore,

$\Rightarrow \quad 2{(2)^3} - k{(2)^2} + (5k - 3) \times 2 - 8 = 14$

$\Rightarrow \quad 16 - 4k + 10k - 6 - 8 = 14$

$\Rightarrow \quad 6k = 12{\text{ and }}k = 2$

4. The polynomials $3{x^3} - a{x^2} + 5x - 13$ and $(a + 1){x^2} - 7x + 5$ leave the same remainder when divided by $x - 3$. Find the value of ' $a$ '.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

$x - 3 = 0 \Rightarrow x = 3$

Since, the given polynomials leave the same remainder when divided by $x - 3$.

Value of polynomial $3{x^3} - a{x^2} + 5x - 13$ at $x = 3$ is the same as the value of polynomial $(a + 1){x^2} - 7x + 5$ at $x = 3$

$\Rightarrow \quad 3{(3)^3} - a{(3)^2} + 5 \times 3 - 13 = (a + 1){(3)^2} - 7 \times 3 + 5$

$\Rightarrow \quad 81 - 9a + 15 - 13 = 9a + 9 - 21 + 5$

$\Rightarrow \quad 18a = 90{\text{ and }}a = 5$

5. When $f(x) = {x^3} + a{x^2} - bx - 8$ is divided by $x - 2$, the remainder is zero and when divided by $x + 1$, the remainder is $ - 30$. Find the values of ' $a$ ' and ' $b$ '.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Since, $x - 2 = 0 \Rightarrow x = 2$

And, given that on dividing $f(x) = {x^3} + a{x^2} - bx - 8$ by $x - 2$, the remainder $= 0$

Then, $f(2) = 0$

$\Rightarrow $ ${2^3} + a{(2)^2} - b(2) - 8 = 0$

$\Rightarrow 8 + 4a - 2b - 8 = 0$

$\Rightarrow 4a - 2b = 0$

$\Rightarrow 2a - b = 0 \to (1)$

Again, given that on dividing $f(x) = {x^3} + a{x^2} - bx - 8$ by $x + 1$, the remainder $=  - 30$

$\Rightarrow f( - 1) =  - 30$

$\Rightarrow {( - 1)^3} + a{( - 1)^2} - b( - 1) - 8 =  - 30$

$\Rightarrow  - 1 + a + b - 8 =  - 30$

$\Rightarrow a + b =  - 21 \to (2)$

When adding $(1)$ and $(2)$, we get  $a =  - 7$.

When substituting the value of $a$ in $(2)$, we get $b =  - 14$.

6. What number should be added to $2{x^3} - 3{x^2} + x$ so that when the resulting polynomial is divided by $x - 2$, the remainder is $3$?

Ans:

Let the number added be $k$ so the resulting polynomial is,

$2{x^3} - 3{x^2} + x + k$

Here, $x - 2 = 0 \Rightarrow x = 2$

Given, when this polynomial is divided by $x - 2$, the remainder $= 3$

$\Rightarrow $$2{(2)^3} - 3{(2)^2} + 2 + k = 3$

Substitute the value of $x$and simplify,

$\Rightarrow 16 - 12 + 2 + k = 3$

$\Rightarrow k =  - 3$

$\therefore $ The required number to be added is $ - 3$.

7. Determine whether $x - 1$ is a factor of ${x^6} - {x^5} + {x^4} + {x^3} - {x^2} - x + 1$ or not ?

Ans:

Since, $x - 1 = 0 \Rightarrow x = 1$

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

When given polynomial is divided by $x - 1$, the remainder is,

$= {(1)^6} - {(1)^5} + {(1)^4} + {(1)^3} - {(1)^2} - (1) + 1$

$= 1 - 1 + 1 + 1 - 1 - 1 + 1$

$= 4 - 3 = 1$, which is not equal to zero.

Therefore, $x - 1$ is not a factor of the given polynomial.

8. If $x - 2$ is a factor of ${x^2} - 7x + 2a$, find the value of $a$.

Ans:

Here, $x - 2 = 0 \Rightarrow x = 2$

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Since, $x - 2$ is a factor of polynomial ${x^2} - 7x + 2a$

$\Rightarrow {\text{ Remainder }} = 0 \Rightarrow {(2)^2} - 7(2) + 2a = 0$$

$\Rightarrow a = 5$

Therefore, the value of $a$is $5.$

9. Find the value of ‘$k$’ if $(x - 2)$ is a factor of ${x^3} + 2{x^2} - kx + 10$. Hence, determine whether $(x + 5)$ is also a factor.

Ans:

Here, $x - 2$ is a factor and $x - 2 = 0 \Rightarrow x = 2$

Thus, the value of given expression ${x^3} + 2{x^2} - kx + 10$ is zero at $x = 2$.

That is, remainder $= 0$

$\Rightarrow {(2)^3} + 2{(2)^2} - k \times 2 + 10{\text{ }} = 0$

$\Rightarrow 8 + 8 - 2k + 10 = 0$

$\Rightarrow k = 13$

On substituting $k = 13$, the given expression becomes ${x^3} + 2{x^2} - 13x + 10$

Now to check whether $(x + 5)$ is also a factor or not,

Find the value of the given expression for $x =  - 5.\quad [\because x + 5 = 0 \Rightarrow x =  - 5]$

${x^3} + 2{x^2} - 13x + 10({\text{ at }}x =  - 5)$

$= {( - 5)^3} + 2{( - 5)^2} - 13( - 5) + 10$

$=  - 125 + 50 + 65 + 10$

$=  - 125 + 125 = 0$

Since, the remainder is $0,(x + 5)$ is a factor.

10. Given that $x + 2$ and $x - 3$ are factors of ${x^3} + ax + b$; calculate the values of $a$ and $b$.

Ans:

Given, $x + 2$ is a factor of ${x^3} + ax + b$

$\Rightarrow {( - 2)^3} + a( - 2) + b = 0[x + 2 = 0 \Rightarrow x =  - 2]$

$\Rightarrow  - 2a + b = 8 \to (1)$

Again, given that,

$x - 3$ is a factor of ${x^3} + ax + b$

$x - 3$ is a factor of ${x^3} + ax + b$

$\Rightarrow $${(3)^3} + a(3) + b = 0$   $[x - 3 = 0 \Rightarrow x = 3]$

$\Rightarrow $$3a + b =  - 27 \to (2)$

On solving equations $(1)$ and $(2)$, we get $a =  - 7$ and $b =  - 6$.

11. Polynomial ${x^3} - a{x^2} + bx - 6$ leaves remainder $ - 8$ when divided by $x - 1$ and $x - 2$ is a factor of it. Find the values of ' $a$ ' and ' $b$ '.

Ans:

On dividing by $x - 1$, the polynomial ${x^3} - b{x^2} + bx - 6$ leaves the remainder $ - 8$.

$\Rightarrow {(1)^3} - a{(1)^2} + b(1) - 6 =  - 8\quad [x - 1 = 0 \Rightarrow x = 1]$

$\Rightarrow $$ - a + b =  - 3$

$\Rightarrow a - b = 3 \to (1)$

$(x - 2)$ is a factor of polynomial ${x^3} - b{x^2} + bx - 6$.

$\Rightarrow {(2)^3} - a{(2)^2} + b(2) - 6 = 0$

$\Rightarrow 8 - 4a + 2b - 6 = 0$

$\Rightarrow 2a - b = 1 \to (2)$

When subtracting $(1)$ and $(2)$, we get $a =  - 2$.

When substituting the value of $a$ in $(1)$, we get  $b =  - 54$.

12. Using the Factor Theorem, show that $(x - 2)$ is a factor of $3{x^2} - 5x - 2$ Hence, factorise the given expression.

Ans:

Since, $x - 2 = 0 \Rightarrow x = 2$

$\therefore $ Remainder $= $ The value of $3{x^2} - 5x - 2$ at $x = 2$.

$= 3{(2)^2} - 5(2) - 2$

$= 12 - 10 - 2 = 0$

$\Rightarrow (x - 2)$ is a factor of $3{x^2} - 5x - 2$.

Now, divide $\left( {3{x^2} - 5x - 2} \right)$ by $(x - 2)$,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x+1\\ x-2)\overline{3x^2- 5x-2} \\\;\;\;\underline{ 3x^2-6x}\\x-2\\\underline{x-2}\\0$

The quotient $= 3x + 1$.

Therefore, $3{x^2} - 5x - 2 = (x - 2)(3x + 1)$.

13. Show that $2x + 7$ is a factor of $2{x^3} + 5{x^2} - 11x - 14.$ Hence, factorise the given expression completely, using the factor theorem.

Ans:

$2x + 7 = 0 \Rightarrow x =  - \frac{7}{2}$

Remainder $= {\text{ Value of }}2{x^3} + 5{x^2} - 11x - 14{\text{ at }}x =  - \frac{7}{2}$

$= 2{\left( { - \frac{7}{2}} \right)^3} + 5{\left( { - \frac{7}{2}} \right)^2} - 11\left( { - \frac{7}{2}} \right) - 14$

$=  - \frac{{343}}{4} + \frac{{245}}{4} + \frac{{77}}{2} - 14$

$= \frac{{ - 343 + 245 + 154 - 56}}{4}$

$= 0$

$ \Rightarrow (2x + 7){\text{  is a factor of }}2{x^3} + 5{x^2} - 11x - 14$

Now, divide $2{x^3} + 5{x^2} - 11x - 14$ by $2x + 7$,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-x-2\\ 2x+7)\overline{2x^3+5x^2-11x-14} \\ \underline{2x^3+7x^2}\\ {-2x^2-11x-14}\\\underline{-2x^2-7x}\\-14x-14\\\underline{-14x-14}\\ {0} $

Thus, $2{x^3} + 5{x^2} - 11x - 14 = (2x + 7)\left( {{x^2} - x - 2} \right)$

$= (2x + 7)\left( {{x^2} - 2x + x - 2} \right)$

$= (2x + 7)[x(x - 2) + 1(x - 2)]$

$= (2x + 7)(x - 2)(x + 1)$

14. Using the Remainder Theorem, factorise the expression $2{x^3} + {x^2} - 2x - 1$ completely.

Ans:

For $x = 1$, the value of given expression is,

$= 2{(1)^3} + {(1)^2} - 2(1) - 1$

$= 2 + 1 - 2 - 1 = 0$

$\Rightarrow x - 1$ is a factor of $2{x^3} + {x^2} - 2x - 1$.

Now, divide $2{x^3} + {x^2} - 2x - 1$ by $x - 1$,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2x^2+3x+1\\ x-1)\overline{2x^3+x^2-2x-1} \\ \underline{2x^3-2x^2}\\ {3x^2-2x-1}\\\underline{3x^2-3x}\\x-1\\\underline{x-1}\\ {0} $

Thus, $2{x^3} + {x^2} - 2x - 1 = (x - 1)\left( {2{x^2} + 3x + 1} \right)$

$= (x - 1)\left( {2{x^2} + 2x + x + 1} \right)$

$= (x - 1)[2x(x + 1) + 1(x + 1)]$

$= (x - 1)(x + 1)(2x + 1)$

15. Find the values of ' $a$ ' and ' $b$ ' so that the polynomial ${x^3} + a{x^2} + bx - 45$ has $(x - 1)$ and $(x + 5)$ as its factors. For the values of ' $a$ ' and ' $b$ ', as obtained above, factorise the given polynomial completely.

Ans:

$(x - 1)$ is a factor of given polynomial ${x^3} + a{x^2} + bx - 45$

$\Rightarrow $${(1)^3} + a{(1)^2} + b(1) - 45 = 0$  $[x - 1 = 0 \Rightarrow x = 1]$

$ \Rightarrow a + b = 44 \to (1)$

$(x + 5)$ is a factor of given polynomial,

$\Rightarrow {( - 5)^3} + a{( - 5)^2} + b( - 5) - 45 = 0\quad \quad [x + 5 = 0 \Rightarrow x =  - 5]$

$\Rightarrow  - 125 + 25a - 5b - 45 = 0$

$\Rightarrow 5a - b = 34 \to (2)$

When adding $(1)$ and $(2)$, we get $a = 13$.

When substituting the value of $a$ in $(1)$, we get $b = 31.$

$\therefore $ The given polynomial ${x^3} + a{x^2} + bx - 45$ $= {x^3} + 13{x^2} + 31x - 45$

Now divide this polynomial by $(x - 1)$,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2+14x+45\\ x-1)\overline{x^3+13x^2+31x-45} \\ \underline{x^3-x^2}\\ {14x^2+31x-45}\\\underline{14x^2-14x}\\45x-45\\\underline{45x-45}\\ {0} $

Thus, ${x^3} + 13{x^2} + 31x - 45$

$= (x - 1)\left( {{x^2} + 14x + 45} \right)$

$= (x - 1)\left( {{x^2} + 9x + 5x + 45} \right)$

$= (x - 1)[x(x + 9) + 5(x + 9)]$

$= (x - 1)(x + 9)(x + 5)$

16. If $(x - 2)$ is a factor of $2{x^3} - {x^2} - px - 2$

(i) find the value of $p$.

Ans:

$\quad x - 2 = 0 \Rightarrow x = 2$

Since, $(x - 2)$ is a factor of given expression

 $\therefore $ Remainder $= 0$

$ \Rightarrow 2{(2)^3} - {(2)^2} - p \times 2 - 2 = 0$

$ \Rightarrow 10 - 2p = 0$

$ \Rightarrow p = 5$

(ii) with the value of $p$, factorise the above expression completely.

Ans:

Substitute the value of $p$in given polynomial,

$2{x^3} - {x^2} - px - 2 = 2{x^3} - {x^2} - 5x - 2$

Now, divide $2{x^3} - {x^2} - 5x - 2$ by $x - 2$,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2x^2+3x+1\\ x-2)\overline{2x^3-x^2-5x-2} \\ \underline{2x^3-4x^2}\\ {3x^2-5x-2}\\\underline{3x^2-6x}\\x-2\\\underline{x-2}\\ {0} $

Thus, $2{x^3} - {x^2} - 5x - 2$

$= (x - 2)\left( {2{x^2} + 3x + 1} \right)$

$= (x - 2)\left( {2{x^2} + 2x + x + 1} \right)$

$= (x - 2)[2x(x + 1) + 1(x + 1)]$

$= (x - 2)(x + 1)(2x + 1)$.

EXERCISE 8(A)

1. Find, in each case, the remainder when:

(i) ${x^4} - 3{x^2} + 2x + 1$ is divided by $x - 1$.

Ans:

By the remainder theorem we know that when a polynomial $f(x)$ is divided by $x - a$, then the remainder is $f(a)$.

Here, $f(x) = {x^4} - 3{x^2} + 2x + 1$

Remainder$= f(1) = {(1)^4} - 3{(1)^2} + 2(1) + 1$

$= 1 - 3 + 2 + 1$

$= 1$

(ii) ${x^3} + 3{x^2} - 12x + 4$ is divided by $x - 2$.

Ans:

By the remainder theorem we know that when a polynomial $f(x)$ is divided by $x - a$, then the remainder is $f(a)$.

Here, $f(x) = {x^3} + 3{x^2} - 12x + 4$

Remainder$= f(2) = {(2)^3} + 3{(2)^2} - 12(2) + 4$

$= 8 + 12 - 24 + 4$

$= 0$

(iii) ${x^4} + 1$ is divided by $x + 1$.

Ans:

By the remainder theorem we know that when a polynomial $f(x)$ is divided by $x - a$, then the remainder is $f(a)$.

Here, $f(x) = {x^4} + 1$

Remainder$= f( - 1) = {( - 1)^4} + 1$

$= 1 + 1$

$= 2$

2. Show that:

(i) $x - 2$ is a factor of $5{x^2} + 15x - 50$

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Then, $x - 2 = 0$

$x = 2$

Remainder$= f(2)$

$= 5{(2)^2} + 15(2) - 50$

$= 0$

The remainder of the polynomial $f(x) = 5{x^2} + 15x - 50$  is $0$ .

Therefore,  $x - 2$  is a factor of polynomial $f(x) = 5{x^2} + 15x - 50$ .

(ii) $3x + 2$ is a factor of $3{x^2} - x - 2$.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Then, $3x + 2 = 0$

$x =  - \frac{2}{3}$

Remainder$= f\left( { - \frac{2}{3}} \right)$

$= 3{\left( { - \frac{2}{3}} \right)^2} - \left( { - \frac{2}{3}} \right) - 2$

$= 0$

The remainder of the polynomial $3{x^2} - x - 2$  is $0$ .

Therefore,  $3x + 2$  is a factor of polynomial $3{x^2} - x - 2$ .

3. Use the Remainder Theorem to find which of the following is a factor of $2{x^3} + 3{x^2} - 5x - 6$.

(i) $x + 1$

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Then, $x + 1 = 0$

$x =  - 1$

Remainder$= f( - 1)$

$= 2{( - 1)^3} + 3{( - 1)^2} - 5( - 1) - 6$

$= 0$

The remainder of the polynomial $f(x) = 2{x^3} + 3{x^2} - 5x - 6$ is $0$ .

Therefore, $x + 1$ is a factor of polynomial $f(x) = 2{x^3} + 3{x^2} - 5x - 6$.

(ii) $2x - 1$

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Then, $2x - 1 = 0$

$x = \frac{1}{2}$

Remainder $= f\left( {\frac{1}{2}} \right)$ 

$= 2{\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} - 5\left( {\frac{1}{2}} \right) - 6$

$=  - \frac{{15}}{2}$

The remainder of the polynomial $f(x) = 2{x^3} + 3{x^2} - 5x - 6$ is $ - \frac{{15}}{2} \ne 0.$

Therefore, $2x - 1$ is a factor of polynomials $f(x) = 2{x^3} + 3{x^2} - 5x - 6$.

(iii) $x + 2$

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Then, Then, $x + 2 = 0$

$x =  - 2$

Remainder$= f( - 2)$

$= 2{( - 2)^3} + 3{( - 2)^2} - 5( - 2) - 6$

$= 0$

The remainder of the polynomial $f(x) = 2{x^3} + 3{x^2} - 5x - 6$ is $0$.

Therefore, $x + 2$ is a factor of polynomials $f(x) = 2{x^3} + 3{x^2} - 5x - 6$.

4. (i) If $2x + 1$ is a factor of $2{x^2} + ax - 3$, find the value of $a$.

Ans:

Here, $2x + 1$ is a factor of $f(x) = 2{x^2} + ax - 3$.

Then, $f\left( {\frac{{ - 1}}{2}} \right) = 0$

$ \Rightarrow 2{\left( {\frac{{ - 1}}{2}} \right)^2} + a\left( {\frac{{ - 1}}{2}} \right) - 3 = 0$

$ \Rightarrow \frac{1}{2} - \frac{a}{2} = 3$

$ \Rightarrow 1 - a = 6$

$ \Rightarrow a =  - 5$

(ii) Find the value of $k$, if $3x - 4$ is a factor of expression $3{x^2} + 2x - k$.

Ans:

Here, $3x - 4$ is a factor of $f(x) = 3{x^2} + 2x - k$.

Then, $f\left( {\frac{4}{3}} \right) = 0$

$ \Rightarrow 3{\left( {\frac{4}{3}} \right)^2} + 2\left( {\frac{4}{3}} \right) - k = 0$

$ \Rightarrow \frac{{16}}{3} + \frac{8}{3} - k = 0$

$ \Rightarrow \frac{{24}}{3} = k$

$ \Rightarrow k = 8$

5. Find the values of constants $a$ and $b$ when $x - 2$ and $x + 3$ both are the factors of expression ${x^3} + a{x^2} + bx - 12$.

Ans:

Let $f(x) = {x^3} + a{x^2} + bx - 12$ 

$x - 2$ is a factor of $f(x)$. 

Then, $x = 2$

Now, remainder $= 0$

${(2)^3} + a{(2)^2} + b(2) - 12 = 0$

$ \Rightarrow 8 + 4a + 2b - 12 = 0$

$ \Rightarrow 4a + 2b - 4 = 0$

$ \Rightarrow 2a + b - 2 = 0 \to (1)$

$x + 3$ is a factor of $f(x)$.

Then, $x =  - 3$

Now, remainder $= 0$

${( - 3)^3} + a{( - 3)^2} + b( - 3) - 12 = 0$

$ \Rightarrow  - 27 + 9a - 3b - 12 = 0$

$ \Rightarrow 9a - 3b - 39 = 0$

$ \Rightarrow 3a - b - 13 = 0 \to (2)$

Adding \[ \left( 1 \right)\] and$\left( 2 \right)$, we get,

$5a - 15 = 0$

$ \Rightarrow a = 3$

Putting the value of a in \[ \left( 1 \right)\], we get,

$6 + b - 2 = 0$

$ \Rightarrow b =  - 4$

Thus, the value of constant $a$ is $3$ and $b$is $ - 4.$

6. Find the value of $k$, if $2x + 1$ is a factor of $(3k + 2){x^3} + (k - 1)$.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Let $f(x) = (3k + 2){x^3} + (k - 1)$

$2x + 1 = 0 \Rightarrow x = \frac{{ - 1}}{2}$

Since, $2x + 1$ is a factor of $f(x)$, remainder is $0$.

Then, $(3k + 2){\left( {\frac{{ - 1}}{2}} \right)^3} + (k - 1) = 0$

$ \Rightarrow \frac{{ - (3k + 2)}}{8} + (k - 1) = 0$

$ \Rightarrow \frac{{ - 3k - 2 + 8k - 8}}{8} = 0$

$ \Rightarrow 5k - 10 = 0$

$ \Rightarrow k = 2$

7. Find the value of $a$, if $x - 2$ is a factor of $2{x^5} - 6{x^4} - 2a{x^3} + 6a{x^2} + 4ax + 8$.

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Let$f(x) = 2{x^5} - 6{x^4} - 2a{x^3} + 6a{x^2} + 4ax + 8$ 

$x - 2 = 0 \Rightarrow x = 2$ 

Since, $x - 2$ is a factor of $f(x)$, remainder $= 0$ 

Then, $2{(2)^5} - 6{(2)^4} - 2a{(2)^3} + 6a{(2)^2} + 4a(2) + 8 = 0$ 

$64 - 96 - 16a + 24a + 8a + 8 = 0$ 

$ - 24 + 16a = 0$ 

$16a = 24$ 

$a = 1.5$

8. Find the values of $m$ and $n$ so that $x - 1$ and $x + 2$ both are factors of ${x^3} + \left( {3m + 1} \right){x^2} + nx - 18.$

Ans:

When a polynomial $f(x)$ is divided by $x - a$, the remainder $= f(a)$. And if the remainder $f(a) = 0;x - a$ is a factor of the polynomial $f(x)$.

Let $f(x) = {x^3} + (3m + 1){x^2} + nx - 18$ 

$x - 1 = 0 \Rightarrow x = 1$ 

$x - 1$ is a factor of $f(x)$. 

So, remainder $= 0$ 

Now, ${(1)^3} + (3m + 1){(1)^2} + n(1) - 18 = 0$ 

$ \Rightarrow 1 + 3m + 1 + n - 18 = 0$ 

$ \Rightarrow 3m + n - 16 = 0 \to (1)$ 

$x + 2 = 0 \Rightarrow x =  - 2$ 

$x + 2$ is a factor of $f(x)$. 

So, remainder $= 0$ 

Now, ${( - 2)^3} + (3m + 1){( - 2)^2} + n( - 2) - 18 = 0$ 

$ \Rightarrow  - 8 + 12m + 4 - 2n - 18 = 0$ 

$ \Rightarrow 12m - 2n - 22 = 0$ 

$ \Rightarrow 6m - n - 11 = 0 \to (2)$ 

Adding $(1)$ and $(2)$, we get, 

$ \Rightarrow 9m - 27 = 0$ 

$ \Rightarrow m = 3$ 

Putting the value of $m$ in $(1)$, we get, 

$3(3) + n - 16 = 0$ 

$ \Rightarrow 9 + n - 16 = 0$ 

$ \Rightarrow n = 7$

9. When ${x^3} + 2{x^2} - kx + 4$ is divided by $x - 2$, the remainder is $k$. Find the value of constant $k$.

Ans:

Let $f(x) = {x^3} + 2{x^2} - kx + 4$ $x - 2 = 0 \Rightarrow x = 2$

On dividing $f(x)$ by $x - 2$, it leaves a remainder $k$.

$f(2) = k$

$ \Rightarrow {(2)^3} + 2{(2)^2} - k(2) + 4 = k$

$ \Rightarrow 8 + 8 - 2k + 4 = k$

$ \Rightarrow 20 = 3k$

$ \Rightarrow k = \frac{{20}}{3} = 6\frac{2}{3}$

10. Find the value of $a$, if the division of $a{x^3} + 9{x^2} + 4x - 10$ by $x + 3$leaves a remainder $5$.

Ans:

Let $f(x) = a{x^3} + 9{x^2} + 4x - 10$

$x + 3 = 0 \Rightarrow x =  - 3$

On dividing $f(x)$ by $x + 3$, it leaves a remainder $5$.

$f( - 3) = 5$

$ \Rightarrow a{( - 3)^3} + 9{( - 3)^2} + 4( - 3) - 10 = 5$

$ \Rightarrow  - 27a + 81 - 12 - 10 = 5$

$ \Rightarrow 54 = 27a$

$ \Rightarrow a = 2$

11. If ${x^3} + a{x^2} + bx + 6$has $x - 2$ as a factor and leaves a remainder $3$ when divided by$x - 3$, find the values of $a$ and $b$.

Ans:

Let $f(x) = {x^3} + a{x^2} + bx + 6$

$x - 2 = 0 \Rightarrow x = 2$

Since, $x - 2$ is a factor, remainder $= 0$

$\therefore f(2) = 0$

${(2)^3} + a{(2)^2} + b(2) + 6 = 0$

$8 + 4a + b + 6 = 0$

$2a + b + 7 = 0 \to (1)$

$x - 3 = 0 \Rightarrow x = 3$

On dividing $f(x)$ by $x - 3$, it leaves a remainder $3$

$\therefore f(3) = 3$

${(3)^3} + a{(3)^2} + b(3) + 6 = 3$

$27 + 9a + 3b + 6 = 3 \to (2)$

$3a + b + 10 = 0$

Subtracting $(1)$ from $(2)$, we get,

$a + 3 = 0$

$a =  - 3$

Substituting the value of a in $(1)$, we get,

$ - 6 + b + 7 = 0$

$b =  - 1$

12. The expression $2{x^3} + a{x^2} + bx - 2$ leaves remainder ${\mathbf{7}}$ and ${\mathbf{0}}$ when divided by $2x - {\mathbf{3}}$ and $x + {\mathbf{2}}$ respectively. Calculate the values of $a$ and $b$.

Ans:

Let $f(x) = 2{x^3} + a{x^2} + bx - 2$

$$2x - 3 = 0 \Rightarrow x = \frac{3}{2}$$

On dividing $f(x)$ by $2x - 3$, it leaves a remainder $7$

$\therefore 2{\left( {\frac{3}{2}} \right)^3} + a{\left( {\frac{3}{2}} \right)^2} + b\left( {\frac{3}{2}} \right) - 2 = 7$

$\frac{{27}}{4} + \frac{{9a}}{4} + \frac{{3b}}{2} = 9$

$\frac{{27 + 9a + 6b}}{4} = 9$

$27 + 9a + 6b = 36$

$9a + 6b - 9 = 0$

$3a + 2b - 3 = 0 \to (1)$

$x + 2 = 0 \Rightarrow x =  - 2$

On dividing $f(x)$ by $x + 2$, it leaves a remainder $0$.

$\therefore 2{( - 2)^3} + a{( - 2)^2} + b( - 2) - 2 = 0$

$ - 16 + 4a - 2b - 2 = 0$

$4a - 2b - 18 = 0 \to (2)$

Adding $(1)$ and $(2)$, we get,

$7a - 21 = 0$

$a = 3$

Substituting the value of a in $(1)$, we get,

$3(3) + 2b - 3 = 0$

$9 + 2b - 3 = 0$

$2b =  - 6$

$b =  - 3$

13. What number should be added to $3{x^3} - 5{x^2} + 6x$ so that when resulting polynomial is divided by$x - 3$, the remainder is $8$?

Ans:

Let the number $k$ be added and the resulting polynomial be $f(x)$.

So, $f(x) = 3{x^3} - 5{x^2} + 6x + k$

It is given that when $f(x)$ is divided by $(x - 3)$, the remainder is $8$ .

$\therefore f(3) = 8$

$3{(3)^3} - 5{(3)^2} + 6(3) + k = 8$

$81 - 45 + 18 + k = 8$

$54 + k = 8$

$k =  - 46$

Thus, the required number is $ - 46$.

14. What number should be subtracted from ${x^3} + 3{x^2} - 8x + 14$ so that on dividing it with $x - 2$, the remainder is $10$.

Ans:

Let the number to be subtracted be $k$ and the resulting polynomial be $f(x)$.

So, $f(x) = {x^3} + 3{x^2} - 8x + 14 - k$

It is given that when $f(x)$ is divided by $(x - 2)$, the remainder is $10.$

$\therefore f(2) = 10$

${(2)^3} + 3{(2)^2} - 8(2) + 14 - k = 10$

$8 + 12 - 16 + 14 - k = 10$

$18 - k = 10$

$k = 8$

Thus, the required number is $8$.

15. The polynomials $2{x^3} - 7{x^2} + ax - 6$ and ${x^3} - 8{x^2} + \left( {2a + 1} \right)x - 16$ leaves the same remainder when divided by$x - 2$. Find the value of $'a'$.

Ans:

Let $f(x) = 2{x^3} - 7{x^2} + ax - 6$

$x - 2 = 0 \Rightarrow x = 2$

When $f(x)$ is divided by $(x - 2)$, remainder $= f(2)$

$\therefore f(2) = 2{(2)^3} - 7{(2)^2} + a(2) - 6$

$= 16 - 28 + 2a - 6$

$= 2a - 18$

Let $g(x) = {x^3} - 8{x^2} + (2a + 1)x - 16$

When $g(x)$ is divided by $(x - 2)$, remainder $= g(2)$

$\therefore g(2) = {(2)^3} - 2{(2)^2} + (2a + 1)(2) - 16$

$= 8 - 32 + 4a + 2 - 16$

$= 4a - 38$

By the given condition, we have:

$f(2) = g(2)$

$2a - 18 = 4a - 38$

$4a - 2a = 38 - 18$

$2a = 20$

$a = 10$

Thus, the value of $a$ is $10$.

16. If $\left( {x - 2} \right)$ is a factor of the expression $2{x^3} + a{x^2} + bx - 14$ and when the expression is divided by$\left( {x - 3} \right)$, it leaves a remainder $52$, find the values of $a$ and $b$.

Ans:

Since $(x - 2)$ is a factor of polynomial $2{x^3} + a{x^2} + bx - 14$, we have,

$2{(2)^3} + a{(2)^2} + b(2) - 14 = 0$

$ \Rightarrow 16 + 4a + 2b - 14 = 0$

$ \Rightarrow 4a + 2b + 2 = 0$

$ \Rightarrow 2a + b + 1 = 0$

$ \Rightarrow 2a + b =  - 1$

On dividing by $(x - 3)$, the polynomial $2{x^3} + a{x^2} + bx - 14$ leaves the remainder $52$.

$ \Rightarrow 2{(3)^3} + a{(3)^2} + b(3) - 14 = 52$

$ \Rightarrow 54 + 9a + 3b - 14 = 52$

$ \Rightarrow 9a + 3b + 40 = 52$

$ \Rightarrow 9a + 3b = 12$

$ \Rightarrow 3a + b = 4$

Subtracting $(1)$ from $(2)$, we get 

$a = 5$

Substituting $a = 5$ in $(1)$, we get 

$2 \times 5 + b =  - 1$

$ \Rightarrow 10 + b =  - 1$

$ \Rightarrow b =  - 11$

Hence, $a = 5$ and  $b =  - 11$

17. Find $'a'$ if the two polynomials $a{x^3} + 3{x^2} - 9$ and$2{x^3} + 4x + a$, leave the same remainder when divided by$x + 3$.

Ans:

Here, $x + 3 = 0 \Rightarrow x =  - 3$

Since, the given polynomials leave the same remainder when divided by $(x - 3)$, value of polynomial $a{x^3} + 3{x^2} - 9$ at $x =  - 3$ is same as value of polynomial $2{x^3} + 4x + a$ at $x =  - 3$.

$ \Rightarrow a{( - 3)^3} + 3{( - 3)^2} - 9 = 2{( - 3)^3} + 4( - 3) + a$

$ \Rightarrow  - 27a + 27 - 9 =  - 54 - 12 + a$

$ \Rightarrow  - 27a + 18 =  - 66 + a$

$ \Rightarrow 28a = 84$

$ \Rightarrow a = \frac{{84}}{{28}}$

$ \Rightarrow a = 3$

EXERCISE 8(B)

1. Using the Factor Theorem, show that:

(i) $(x - 2)$ is a factor of${x^3} - 2{x^2} - 9x + 18$. Hence, factorise the expression ${x^3} - 2{x^2} - 9x + 18$ completely.

Ans:

Let $f(x) = {x^3} - 2{x^2} - 9x + 18$ 

$x - 2 = 0 \Rightarrow x = 2$

$\therefore $ Remainder $= f(2)$ 

$= {(2)^3} - 2{(2)^2} - 9(2) + 18$ 

$= 8 - 8 - 18 + 18$ 

$= 0$

Hence, $(x - 2)$ is a factor of $f(x)$.

Now, we have:

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-9\\ x-2)\overline{x^3-2x^2-9x+18} \\ \underline{x^3-2x^2}\\ {-9x+18}\\\underline{-9x+18}\\ {0} $

$\therefore {x^3} - 2{x^2} - 9x + 18 = (x - 2)\left( {{x^2} - 9} \right) = (x - 2)(x + 3)(x - 3)$.

(ii) $(x + 5)$ is a factor of $2{x^3} + 5{x^2} - 28x - 15$. Hence, factorise the expression $2{x^3} + 5{x^2} - 28x - 15$ completely.

Ans:

Let $f(x) = 2{x^3} + 5{x^2} - 28x - 15$

$x + 5 = 0 \Rightarrow x =  - 5$

$\therefore $ Remainder $= {\text{f}}( - 5)$

$= 2{( - 5)^3} + 5{( - 5)^2} - 28( - 5) - 15$

$=  - 250 + 125 + 140 - 15$

$=  - 265 + 265$

$= 0$

Hence, $(x + 5)$ is a factor of $f(x)$.

Now, we have:

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-9\\ x-2)\overline{x^3-2x^2-9x+18} \\ \underline{x^3-2x^2}\\ {-9x+18}\\\underline{-9x+18}\\x-2\\\underline{x-2}\\ {0} $

$\therefore 2{x^3} + 5{x^2} - 28x - 15 = (x + 5)\left( {2{x^2} - 5x - 3} \right)$

$= (x + 5)\left[ {2{x^2} - 6x + x - 3} \right]$

$= (x + 5)[2x(x - 3) + 1(x - 3)]$

$= (x + 5)(2x + 1)(x - 3)$

(iii) $(3x + 2)$ is a factor of $3{x^3} + 2{x^2} - 3x - 2$. Hence, factorise the expression $3{x^3} + 2{x^2} - 3x - 2$ completely.

Ans:

Let $f(x) = 3{x^3} + 2{x^2} - 3x - 2$

$3x + 2 = 0 \Rightarrow x = \frac{{ - 2}}{3}$

$\therefore $ Remainder $= f\left( {\frac{{ - 2}}{3}} \right)$

$= 3{\left( {\frac{{ - 2}}{3}} \right)^3} + 2{\left( {\frac{{ - 2}}{3}} \right)^2} - 3\left( {\frac{{ - 2}}{3}} \right) - 2$

$= \frac{{ - 8}}{9} + \frac{8}{9} + 2 - 2$

$= 0$

Hence, $(3x + 2)$ is a factor of $f(x)$.

Now, we have:

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-1\\ 3x+2)\overline{3x^3+2x^2-3x-2} \\ \underline{3x^3+2x^2}\\ {-3x-2}\\\underline{-3x-2}\\ {0} $

$\therefore 3{x^3} + 2{x^2} - 3x - 2 = (3x + 2)\left( {{x^2} - 1} \right) = (3x + 2)(x + 1)(x - 1)$.

2. Using the Remainder Theorem, factorise each of the following completely.

(i) $3{x^3} + 2{x^2} - 19x + 6$

Ans:

For $x = 2$, the value of the given expression $3{x^3} + 2{x^2} - 19x + 6$

$f(2) = 3{(2)^3} + 2{(2)^2} - 19(2) + 6$

$= 24 + 8 - 38 + 6$

$ \Rightarrow x - 2{\text{ is a factor of }}3{x^3} + 2{x^2} - 19x + 6$

Now let us do long division.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x^2+8x-3\\ x-2)\overline{3x^3+2x^2-19x+6} \\ \underline{3x^3-6x^2}\\ {8x^2-19x}\\\underline{8x^2-16x}\\{-3x+16}\\\underline{-3x+16}\\ {0} $

Thus, we have,

$3{x^3} + 2{x^2} - 19x + 6 = (x - 2)\left( {3{x^2} + 8x - 3} \right)$

$= (x - 2)\left( {3{x^2} + 9x - x - 3} \right)$

$= (x - 2)(3x(x + 3) - (x + 3))$

$= (x - 2)(3x - 1)(x + 3)$

(ii) $2{x^3} + {x^2} - 13x + 6$

Ans:

Let $f(x) = 2{x^3} + {x^2} - 13x + 6$

For $x = 2$,

$f(2) = 2{(2)^3} + {(2)^2} - 13(2) + 6$

$= 16 + 4 - 26 + 6$

$= 0$

Hence, $(x - 2)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2x^2+5x-3\\ x-2)\overline{2x^3+x^2-13x+6} \\ \underline{2x^3-4x^2}\\ {5x^2-13x}\\\underline{5x^2-10x}\\{-3x+16}\\\underline{-3x+16}\\ {0} $

Now, $2{x^3} + {x^2} - 13x + 6 = (x - 2)\left( {2{x^2} + 5x - 3} \right)$

$= (x - 2)\left( {2{x^2} + 6x - x - 3} \right)$

$= (x - 2)[2x(x + 3) - (x + 3)]$

$= (x - 2)(x + 3)(2x - 1)$

(iii) $3{x^3} + 2{x^2} - 23x - 30$

Ans:

Let $f(x) = 3{x^3} + 2{x^2} - 23x - 30$

For $x =  - 2$,

$f( - 2) = 3{( - 2)^3} + 2{( - 2)^2} - 23( - 2) - 30$

$=  - 24 + 8 + 46 - 30$

$=  - 54 + 54$

$= 0$

Hence, $(x + 2)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x^2-4x-15\\ x+2)\overline{3x^3+2x^2-23x-30} \\ \underline{3x^3+6x^2}\\ {-4x^2-23x}\\\underline{-4x^2-8x}\\{-15x-30}\\\underline{-15x-30}\\ {0} $

Now, $3{x^3} + 2{x^2} - 23x - 30 = (x + 2)\left( {3{x^2} - 4x - 15} \right)$

$= (x + 2)\left( {3{x^2} + 5x - 9x - 15} \right)$

$= (x + 2)[x(3x + 5) - 3(3x + 5)]$

$= (x + 2)(3x + 5)(x - 3)$

(iv) $4{x^3} + 7{x^2} - 36x - 63$

Ans:

Let $f(x) = 4{x^3} + 7{x^2} - 36x - 63$

For $x = 3$,

$f(3) = 4{(3)^3} + 7{(3)^2} - 36(3) - 63$

$= 108 + 63 - 108 - 63$

$= 0$

Hence, $(x + 3)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4x^2-5x-21\\ (x+3))\overline{4x^3+7x^2-36x-63} \\ \underline{4x^3+2x^2}\\ {-5x^2-36x}\\\underline{-5x^2-15x}\\{-21x-63}\\\underline{-21x-63}\\ {0} $

Now, $4{x^3} + 7{x^2} - 36x - 63 = (x + 3)\left( {4{x^2} - 5x - 21} \right)$

$= (x + 3)\left( {4{x^2} - 12x + 7x - 21} \right)$

$= (x + 3)[4x(x - 3) + 7(x - 3)]$

$= (x + 3)(4x + 7)(x - 3)$

(v) ${x^3} + {x^2} - 4x - 4$

Ans:

Let $f(x) = {x^3} + {x^2} - 4x - 4$

For $x =  - 1$,

$f( - 1) = {( - 1)^3} + {( - 1)^2} - 4( - 1) - 4$

$=  - 1 + 1 + 4 - 4$

$= 0$

Hence, $(x + 1)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-4\\ (x+1))\overline{x^3+x^2-4x-4} \\ \underline{x^3+x^2}\\ {-4x-4}\\\underline{-4x-4} \\ {0} $

Now, ${x^3} + {x^2} - 4x - 4$

$= (x + 1)\left( {{x^2} - 4} \right)$

$= (x + 1)(x + 2)(x - 2)$

3. Using the Remainder Theorem, factorise the expression$3{x^3} + 10{x^2} + x - 6$. Hence, solve the equation $3{x^3} + 10{x^2} + x - 6 = 0$.

Ans:

Let $f(x) = 3{x^3} + 10{x^2} + x - 6$

For $x =  - 1$,

$f( - 1) = 3{( - 1)^3} + 10{( - 1)^2} + ( - 1) - 6$

$=  - 3 + 10 - 1 - 6$

$= 0$

Hence, $(x + 1)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x^2+7x-6\\ (x+1))\overline{3x^3+10x^2+x-6} \\ \underline{3x^3+3x^2}\\ {7x^2+x}\\\underline{7x^2+7x}\\{-6x-6}\\\underline{-6x-6}\\ {0} $

$\therefore 3{x^3} + 10{x^2} + x - 6 = (x + 1)\left( {3{x^2} + 7x - 6} \right)$

$= (x + 1)\left( {3{x^2} + 9x - 2x - 6} \right)$

$= (x + 1)[3x(x + 3) - 2(x + 3)]$

$= (x + 1)(x + 3)(3x - 2)$

Now, $3{x^3} + 10{x^2} + x - 6 = 0$

$ \Rightarrow (x + 1)(x + 3)(3x - 2) = 0$

$ \Rightarrow x =  - 1, - 3,\frac{2}{3}$

4. Factorise the expression$f\left( x \right) = 2{x^3} - 7{x^2} - 3x + 18$. Hence, find all possible values of $x$ for which$f\left( x \right) = 0$.

Ans:

Let $f(x) = 2{x^3} - 7{x^2} - 3x + 18$

For $x = 2$,

$f(2) = 2{(2)^3} - 7{(2)^2} - 3(2) + 18$

$= 16 - 28 - 6 + 18$

$= 0$

Hence, $(x - 2)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2x^2-3x-9\\ (x-2))\overline{2x^3-7x^2-3x+18} \\ \underline{2x^3-4x^2}\\ {-3x^2-3x}\\\underline{-3x^2+6x}\\{-9x+18}\\\underline{-9x+18}\\ {0} $

$\therefore 2{x^3} - 7{x^2} - 3x + 18 = (x - 2)\left( {2{x^2} - 3x - 9} \right)$

$= (x - 2)\left( {2{x^2} - 6x + 3x - 9} \right)$

$= (x - 2)[2x(x - 3) + 3(x - 3)]$

$= (x - 2)(x - 3)(2x + 3)$

Now, $f(x) = 0$

$ \Rightarrow 2{x^3} - 7{x^2} - 3x + 18 = 0$

$ \Rightarrow (x - 2)(x - 3)(2x + 3) = 0$

$ \Rightarrow x = 2,3,\frac{{ - 3}}{2}$

5. Given that $x - 2$ and $x + 1$ are factors of$f\left( x \right) = {x^3} + 3{x^2} + ax + b$ ; calculate the values of $a$ and $b$. Hence, find all the factors of $f\left( x \right).$

Ans:

$f(x) = {x^3} + 3{x^2} + ax + b$

Since, $(x - 2)$ is a factor of $f(x),f(2) = 0$

$ \Rightarrow {(2)^3} + 3{(2)^2} + a(2) + b = 0$

$ \Rightarrow 8 + 12 + 2a + b = 0$

$ \Rightarrow 2a + b + 20 = 0 \to (1)$

Since, $(x + 1)$ is a factor of $f(x),f( - 1) = 0$

$ \Rightarrow {( - 1)^3} + 3{( - 1)^2} + a( - 1) + b = 0$

$ \Rightarrow  - 1 + 3 - a + b = 0$

$ \Rightarrow  - a + b + 2 = 0 \to (2)$

Subtracting $(2)$ from $(1)$, we get,$3a + 18 = 0$

$ \Rightarrow a =  - 6$

Substituting the value of $a$ in $(2)$, we get,

$b = a - 2 =  - 6 - 2 =  - 8$

$\therefore f(x) = {x^3} + 3{x^2} - 6x - 8$

Now, for $x =  - 1$,

 $f( - 1) = {( - 1)^3} + 3{( - 1)^2} - 6( - 1) - 8$

$=  - 1 + 3 + 6 - 8$

$= 0$

Hence, $(x + 1)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2+2x-8\\ (x+1))\overline{x^3+3x^2-6x-8} \\ \underline{x^2+x^2}\\ {-2x^2-6x}\\\underline{2x^2+2x}\\{-8x-8}\\\underline{-8x-8}\\ {0} $

$\therefore {x^3} + 3{x^2} - 6x - 8 = (x + 1)\left( {{x^2} + 2x - 8} \right)$

$= (x + 1)\left( {{x^2} + 4x - 2x - 8} \right)$

$= (x + 1)[x(x + 4) - 2(x + 4)]$

$= (x + 1)(x + 4)(x - 2)$

6. The expression $4{x^3} - b{x^2} + x - c$ leaves remainders $0$ and $30$ when divided by $x + 1$and $2x - 3$ respectively. Calculate the values of $b$ and $c$. Hence, factorise the expression completely.

Ans:

Let $f(x) = 4{x^3} - b{x^2} + x - c$

It is given that when $f(x)$ is divided by $(x + 1)$, the remainder is 0 .

$\therefore f( - 1) = 0$

$4{( - 1)^3} - b{( - 1)^2} + ( - 1) - c = 0$

$ - 4 - b - 1 - c = 0$

$b + c + 5 = 0 \to (1)$

It is given that when $f(x)$ is divided by $(2x - 3)$, the remainder is $30$.

$\therefore f\left( {\frac{3}{2}} \right) = 30$

$4{\left( {\frac{3}{2}} \right)^3} - b{\left( {\frac{3}{2}} \right)^2} + \left( {\frac{3}{2}} \right) - c = 30$

$\frac{{27}}{2} - \frac{{9b}}{4} + \frac{3}{2} - c = 30$

$54 - 9b + 6 - 4c - 120 = 0$

$9b + 4c - 60 = 0 \to (2)$

Multiplying $(1)$ by $4$ and subtracting it from $(2)$, we get,

$5b + 40 = 0$

$b =  - 8$

Substituting the value of $b$ in $(2)$, we get,

$c =  - 5 + 8 = 3$

Therefore, $f(x) = 4{x^3} + 8{x^2} + x - 3$

Now, for $x =  - 1$, we get,

$f( - 1) = 4{( - 1)^3} + 8{( - 1)^2} + ( - 1) - 3$

$=  - 4 + 8 - 1 - 3$

$= 0$

Hence, $(x + 1)$ is a factor of $f(x)$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4x^2+4x-3\\ (x+1))\overline{4x^3+8x^2+x-3} \\ \underline{4x^3+4x^2}\\ {4x^2+x}\\\underline{4x^2+4x}\\{-3x-3}\\\underline{-3x-3}\\ {0} $

$\therefore 4{x^3} + 8{x^2} + x - 3 = (x + 1)\left( {4{x^2} + 4x - 3} \right)$

$= (x + 1)\left( {4{x^2} + 6x - 2x - 3} \right)$

$= (x + 1)[2x(2x + 3) - (2x + 3)]$

$= (x + 1)(2x + 3)(2x - 1)$

7. If $x + a$ is a common factor of expressions $f\left( x \right) = {x^2} + px + q$ and$g\left( x \right) = {x^2} + mx + n$; show that: $a = \frac{{n - q}}{{m - p}}$.

Ans:

$f(x) = {x^2} + px + q$

It is given that $(x + a)$ is a factor of $f(x)$.

$\therefore f( - a) = 0$

$ \Rightarrow {( - a)^2} + p( - a) + q = 0$

$ \Rightarrow {a^2} - pa + q = 0$

$ \Rightarrow {a^2} = pa - q \to (1)$

$g(x) = {x^2} + mx + n$

It is given that $(x + a)$ is a factor of $g(x)$.

$\therefore g( - a) = 0$

$ \Rightarrow {( - a)^2} + m( - a) + n = 0$

$ \Rightarrow {a^2} - ma + n = 0$

$ \Rightarrow {a^2} = ma - n \to (2)$

From $(1)$ and $(2)$, we get,

$pa - q = ma - n$

$n - q = a(m - p)$

$a = \frac{{n - q}}{{m - p}}$

8. The polynomials $a{x^3} + 3{x^2} - 3$ and$2{x^3} - 5x + a$, when divided by$x - 4$, leave the same remainder in each case. Find the value of $a$.

Ans:
Let $f(x) = a{x^3} + 3{x^2} - 3$

When $f(x)$ is divided by $(x - 4)$, remainder $= f(4)$

$f(4) = a{(4)^3} + 3{(4)^2} - 3 = 64a + 45$

Let $g(x) = 2{x^3} - 5x + a$

When $g(x)$ is divided by $(x - 4)$, remainder $= g(4)$

$g(4) = 2{(4)^3} - 5(4) + a = a + 108$

It is given that $f(4) = g(4)$

$64a + 45 = a + 108$

$63a = 63$

$a = 1$

9. Find the value of $'a'$, if $\left( {x - a} \right)$ is a factor of ${x^3} - a{x^2} + x + 2$.

Ans:

Let $f(x) = {x^3} - a{x^2} + x + 2$

It is given that $(x - a)$ is a factor of $f(x)$.

$\therefore $ Remainder $= {\text{f}}({\text{a}}) = 0$

${a^3} - {a^3} + a + 2 = 0$

$a + 2 = 0$

$a =  - 2$

10. Find the number that must be subtracted from the polynomial $3{y^3} + {y^2} - 22y + 15$, so that the resulting polynomial is completely divisible by $y + 3$.

Ans:

Let the number to be subtracted from the given polynomial be $k$.

Let $f(y) = 3{y^3} + {y^2} - 22y + 15 - k$

It is given that $f(y)$ is divisible by $(y + 3)$.

$\therefore $ Remainder $= f( - 3) = 0$

$3{( - 3)^3} + {( - 3)^2} - 22( - 3) + 15 - k = 0$

$ - 81 + 9 + 66 + 15 - k = 0$

$9 - k = 0$

$k = 9$

EXERCISE 8(C)

1. Show that $\left( {x - 1} \right)$ is a factor of ${x^3} - 7{x^2} + 14x - 8$. Hence, completely factorise the given expression.

Ans:

Let $f(x) = {x^3} - 7{x^2} + 14x - 8$

$f(1) = {(1)^3} - 7{(1)^2} + 14(1) - 8 = 1 - 7 + 14 - 8 = 0$

Hence, $(x - 1)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-6x-8\\ (x-1))\overline{x^3-7x^2+14x-8} \\ \underline{x^3-x^2}\\ {-6x^2+14x}\\\underline{-6x^2+6x}\\{8x-8}\\\underline{8x-8}\\ {0} $

$\therefore {x^3} - 7{x^2} + 14x - 8 = (x - 1)\left( {{x^2} - 6x + 8} \right)$

$= (x - 1)\left( {{x^2} - 2x - 4x + 8} \right)$

$= (x - 1)[x(x - 2) - 4(x - 2)]$

$= (x - 1)(x - 2)(x - 4)$

2. Using Remainder Theorem, factorise:

${x^3} + 10{x^2} - 37x + 26$ completely.

Ans:

By Remainder Theorem,

For $x = 1$, the value of the given expression is the remainder.

${x^3} + 10{x^2} - 37x + 26$

$= {(1)^3} + 10{(1)^2} - 37(1) + 26$

$= 1 + 10 - 37 + 26$

$= 37 - 37$

$= 0$

$ \Rightarrow x - 1{\text{ is a factor of }}{x^3} + 10{x^2} - 37x + 26$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2+11x-26\\ (x-1))\overline{x^3+10x^2-37x+26} \\ \underline{x^3-x^2}\\ {-11x^2-37x}\\\underline{-11x^2-11x}\\{-26x-26}\\\underline{-26x-26}\\ {0} $

$\therefore {x^3} + 10{x^2} - 37x + 26 = (x - 1)\left( {{x^2} + 11x - 26} \right)$

$= (x - 1)\left( {{x^2} + 13x - 2x - 26} \right)$

$= (x - 1)[x(x + 13) - 2(x + 13)]$

$= (x - 1)(x + 13)(x - 2)$

3. When ${x^3} + 3{x^2} - mx + 4$ is divided by$x - 2$, the remainder is$m + 3$. Find the value of$m$.

Ans:

Let $f(x) = {x^3} + 3{x^2} - mx + 4$

According to the given information,

$f(2) = m + 3$

${(2)^3} + 3{(2)^2} - m(2) + 4 = m + 3$

$8 + 12 - 2m + 4 = m + 3$

$24 - 3 = m + 2m$

$3m = 21$

$m = 7$

4. What should be subtracted from$3{x^3} - 8{x^2} + 4x - 3$, so that the resulting expression has $x + 2$ as a factor?

Ans:

Let the required number be $k$.

Let $f(x) = 3{x^3} - 8{x^2} + 4x - 3 - k$

According to the given information,

$f( - 2) = 0$

$3{( - 2)^3} - 8{( - 2)^2} + 4( - 2) - 3 - k = 0$

$ - 24 - 32 - 8 - 3 - k = 0$

$ - 67 - k = 0$

$k =  - 67$

Thus, the required number is $ - 67$.

5. If $(x + 1)$ and $(x - 2)$ are factors of ${x^3} + (a + 1){x^2} - (b - 2)x - 6$, find the values of $a$and $b$. And then, factorise the given expression completely.

Ans:

Let $f(x) = {x^3} + (a + 1){x^2} - (b - 2)x - 6$

Since, $(x + 1)$ is a factor of $f(x)$

$\therefore $ Remainder $= f( - 1) = 0$ 

${( - 1)^3} + (a + 1){( - 1)^2} - (b - 2)( - 1) - 6 = 0$ $ - 1 + (a + 1) + (b - 2) - 6 = 0$ 

$a + b - 8 = 0 \to (1)$

Since, $(x - 2)$ is a factor of $f(x)$

$\therefore $ Remainder $= {\text{f}}(2) = 0$

${(2)^3} + (a + 1){(2)^2} - (b - 2)(2) - 6 = 0$

$8 + 4a + 4 - 2b + 4 - 6 = 0$

$4a - 2b + 10 = 0$

$2a - b + 5 = 0 \to (2)$ 

Substituting the value of a in $({\mathbf{1}})$, we get,

$1 + b - 8 = 0$

$b = 7$

$\therefore f(x) = {x^3} + 2{x^2} - 5x - 6$

Now, $(x + 1)$ and $(x - 2)$ are factors of $f(x)$. 

Hence, $(x + 1)(x - 2) = {x^2} - x - 2$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x+3\\ x^2-x-2)\overline{x^3+2x^2-5x-6} \\ \underline{x^3-x^2-2x}\\ {3x^2-3x-6}\\\underline{3x^2-3x-6}\\ {0} $

$\therefore f(x) = {x^3} + 2{x^2} - 5x - 6 = (x + 1)(x - 2)(x + 3)$

6. If $x - 2$ is a factor of ${x^2} + ax + b$ and $a + b = 1$, find the values of $a$ and $b$.

Ans:

Let $f(x) = {x^2} + ax + b$

Since, $(x - 2)$ is a factor of $f(x)$.

$\therefore $ Remainder $= f(2) = 0$

${(2)^2} + a(2) + b = 0$

$4 + 2a + b = 0$

$2a + b =  - 4 \to (1)$

It is given that,

$a + b = 1 \to (2)$

Subtracting $(2)$ from $(1)$, we get, 

$a =  - 5$

Substituting the value of $a$ in $(2)$, we get, 

$b = 1 - ( - 5) = 6$

7. Factorise ${x^3} + 6{x^2} + 11x + 6$ completely using factor theorem.

Ans:

Let $f(x) = {x^3} + 6{x^2} + 11x + 6$

For $x =  - 1$

$f( - 1) = {( - 1)^3} + 6{( - 1)^2} + 11( - 1) + 6$

$=  - 1 + 6 - 11 + 6$

$= 12 - 12$

$= 0$

Hence, $(x + 1)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2+5x+6\\ (x+1))\overline{x^3+6x^2+11x+6} \\ \underline{x^3+x^2}\\ {5x^2+11x}\\\underline{5x^2+5x}\\{6x+6}\\\underline{6x+6}\\ {0} $

$\therefore {x^3} + 6{x^2} + 11x + 6 = (x + 1)\left( {{x^2} + 5x + 6} \right)$

$= (x + 1)\left( {{x^2} + 2x + 3x + 6} \right)$

$= (x + 1)[x(x + 2) + 3(x + 2)]$

$= (x + 1)(x + 2)(x + 3)$

8. Find the value of $'m'$, if $m{x^3} + 2{x^2} - 3$ and ${x^2} - mx + 4$ leave the same remainder when each is divided by $x - 2$

Ans:

Let $f(x) = m{x^3} + 2{x^2} - 3$

$g(x) = {x^2} - mx + 4$

It is given that $f(x)$ and $g(x)$ leave the same remainder when divided by $(x - 2)$. Therefore, we have:

$f(2) = g(2)$

$m{(2)^3} + 2{(2)^2} - 3 = {(2)^2} - m(2) + 4$

$8m + 8 - 3 = 4 - 2m + 4$

$10m = 3$

$m = \frac{3}{{10}}$

9. The polynomial $p{x^3} + 4{x^2} - 3x + q$ is completely divisible by ${x^2} - 1$; find the values of $p$ and $q$. Also, for these values of $p$ and $q$ factorize the given polynomial completely.

Ans: 

Let $f(x) = p{x^3} + 4{x^2} - 3x + q$

It is given that $f(x)$ is completely divisible by $\left( {{x^2} - 1} \right) = (x + 1)(x - 1)$.

Therefore, $f(1) = 0$ and $f( - 1) = 0$

$f(1) = p{(1)^3} + 4{(1)^2} - 3(1) + q = 0$

$p + q + 1 = 0 \to (1)$

$f( - 1) = p{( - 1)^3} + 4{( - 1)^2} - 3( - 1) + q = 0$ 

$ - p + q + 7 = 0 \to (2)$ 

Adding $(1)$ and $(2)$, we get,

$2q + 8 = 0$

$q =  - 4$

$\therefore f(x) = 3{x^3} + 4{x^2} - 3x - 4$

Given that $f(x)$ is completely divisible by $\left( {{x^2} - 1} \right)$.

Then,

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3x+4\\ (x^2-1))\overline{3x^3+4x^2-3x-4} \\ \underline{3x^3-3x}\\ {4x^2-4}\\\underline{4x^2-4}\\ {0} $

$\therefore 3{x^3} + 4{x^2} - 3x - 4 = \left( {{x^2} - 1} \right)(3x + 4)$

$= (x - 1)(x + 1)(3x + 4)$

10. Find the number which should be added to ${x^2} + x + 3$so that the resulting polynomial is completely divisible by $(x + 3)$.

Ans:

Let the required number be $k$.

Let $f(x) = {x^2} + x + 3 + k$

It is given that $f(x)$ is divisible by $(x + 3)$.

$\therefore $ Remainder $= 0$

$f( - 3) = 0$

${( - 3)^2} + ( - 3) + 3 + k = 0$

$9 - 3 + 3 + k = 0$

$9 + k = 0$

$k =  - 9$

Thus, the required number is $ - 9$.

11. When the polynomial  ${x^3} + 2{x^2} - 5ax - 7$ is divided by $\left( {x - {\mathbf{1}}} \right)$, the remainder is $A$and when the polynomial ${x^3} + a{x^2} - 12x + 16$ is divided by $\left( {x + 2} \right)$, the remainder is  $B$ . Find the value of $'a'$ if  ${\mathbf{2}}A + B = {\mathbf{0}}$.

Ans:

It is given that when the polynomial ${x^3} + 2{x^2} - 5ax - 7$ is divided by $(x - 1)$, the remainder is $A$.

$\therefore {(1)^3} + 2{(1)^2} - 5a(1) - 7 = A$

$1 + 2 - 5a - 7 = A$

$ - 5a - 4 = A \to (1)$

It is also given that when the polynomial ${x^3} + a{x^2} - 12x + 16$ is divided by $(x + 2)$, the remainder is $B$.

$\therefore {x^3} + a{x^2} - 12x + 16 = B$

${( - 2)^3} + a{( - 2)^2} - 12( - 2) + 16 = B$

$ - 8 + 4a + 24 + 16 = B$

$4a + 32 = B \to (2)$

It is also given that $2A + B = 0$

Using $(1)$ and $(2)$, we get,

$2( - 5a - 4) + 4a + 32 = 0$

$ - 10a - 8 + 4a + 32 = 0$

$ - 6a + 24 = 0$

$6a = 24$

$a = 4$

12. $\left( {{\mathbf{3}}x + {\mathbf{5}}} \right)$ is a factor of the polynomial$(a - 1){x^3} + (a + 1){x^2} - (2a + 1)x - 15$. Find the value of $'a'$, factorise the given polynomial completely.

Ans:

Let $f(x) = (a - 1){x^3} + (a + 1){x^2} - (2a + 1)x - 15$

It is given that $(3x + 5)$ is a factor of $f(x)$

$\therefore $ Remainder $= 0$

$f\left( {\frac{{ - 5}}{3}} \right) = 0$

$(a - 1){\left( {\frac{{ - 5}}{3}} \right)^3} + (a + 1){\left( {\frac{{ - 5}}{3}} \right)^2} - (2a + 1)\left( {\frac{{ - 5}}{3}} \right) - 15 = 0$

$(a - 1)\left( {\frac{{ - 125}}{{27}}} \right) + (a + 1)\left( {\frac{{25}}{9}} \right) - (2a + 1)\left( {\frac{{ - 5}}{3}} \right) - 15 = 0$

$\frac{{ - 125(a - 1) + 75(a + 1) + 45(2a + 1) - 405}}{{27}} = 0$

$ - 125a + 125 + 75a + 75 + 90a + 45 - 405 = 0$

$40a - 160 = 0$

$40a = 160$

$a = 4$

$\therefore f(x) = (a - 1){x^3} + (a + 1){x^2} - (2a + 1)x - 15$

$= 3{x^3} + 5{x^2} - 9x - 15$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^2-3\\ 3x+5)\overline{3x^3+5x^2-9x-15} \\ \underline{3x^3+5x^2}\\ {-9x-15}\\\underline{-9x-15}\\ {0} $

$\therefore 3{x^3} + 5{x^2} - 9x - 15 = (3x + 5)\left( {{x^2} - 3} \right)$

$= (3x + 5)\left( {x + \sqrt 3 } \right)\left( {x - \sqrt 3 } \right)$

13. When divided by x-3 the polynomials ${x^3} - p{x^2} + x + 6$ and $2{x^3} - {x^2} - \left( {p + 3} \right)x - 6$ leave the same remainder. Find the value of $'p'$.

Ans:

If $(x - 3)$ divides $f(x) = {x^3} - p{x^2} + x + 6$, then,

Remainder $= f(3) = {3^3} - p{(3)^2} + 3 + 6 = 36 - 9p$

If $(x - 3)$ divides $g(x) = 2{x^3} - {x^2} - (p + 3)x - 6$, then

Remainder $= g(3) = 2{(3)^3} - {(3)^2} - (p + 3)(3) - 6 = 30 - 3p$

Now, $f(3) = g(3)$

$ \Rightarrow 36 - 9p = 30 - 3p$

$ \Rightarrow  - 6p =  - 6$

$ \Rightarrow p = 1$

14. Use the Remainder Theorem to factorise the following expression:

$2{x^3} + {x^2} - 13x + 6$.

Ans:

$f(x) = 2{x^3} + {x^2} - 13x + 6$

Factors of constant term 6 are $ \pm 1, \pm 2, \pm 3, \pm 6$.

Putting $x = 2$, we have:

$f(2) = 2{(2)^3} + {2^2} - 13(2) + 6$

$= 16 + 4 - 26 + 6$

$= 0$

Hence, $(x - 2)$ is a factor of $f(x)$.

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2x^2+5x-3\\ x-2)\overline{2x^3+x^2-13x+6} \\ \underline{2x^3-4x^2}\\ {5x^2-13x}\\\underline{5x^2-10x}\\ {-3x+6}\\\underline{-3x+6}\\{0} $

$2{x^3} + {x^2} - 13x + 6 = (x - 2)\left( {2{x^2} + 5x - 3} \right)$

$= (x - 2)\left( {2{x^2} + 6x - x - 3} \right)$

$= (x - 2)(2x(x + 3) - 1(x + 3))$

$= (x - 2)(2x - 1)(x + 3)$

15. Using the remainder theorem, find the value of $k$ if on dividing $2{x^3} + 3{x^2} - kx + 5$ by$x - 2$, leaves a remainder $7$.

Ans:

Let $f(x) = 2{x^3} + 3{x^2} - kx + 5$

Using Remainder Theorem, we have

$f(2) = 7$

$\therefore 2{(2)^3} + 3{(2)^2} - k(2) + 5 = 7$

$ \Rightarrow 16 + 12 - 2k + 5 = 7$

$ \Rightarrow 33 - 2k = 7$

$ \Rightarrow 2k = 26$

$ \Rightarrow k = 13$

16. What must be subtracted from $16{x^3} - 8{x^2} + 4x + 7$ so that the resulting expression has $2x + 1$ as a factor?

Ans:

Here, $f(x) = 16{x^3} - 8{x^2} + 4x + 7$

Let the number subtracted be $k$ from the given polynomial $f(x)$.

Given that $2x + 1$ is a factor of $f(x)$.

$\therefore f\left( { - \frac{1}{2}} \right) = 0$

$ \Rightarrow 16{\left( { - \frac{1}{2}} \right)^3} - 8{\left( { - \frac{1}{2}} \right)^2} + 4\left( { - \frac{1}{2}} \right) + 7 - k = 0$

$ \Rightarrow 16 \times \left( { - \frac{1}{8}} \right) - 8 \times \frac{1}{4} - 2 + 7 - k = 0$

$ \Rightarrow  - 2 - 2 - 2 + 7 - k = 0$

$ \Rightarrow  - 6 + 7 - k = 0$

$ \Rightarrow k = 1$

Therefore, ${\mathbf{1}}$ must be subtracted from $16{x^3} - 8{x^2} + 4x + 7$ so that the resulting expression has $2x + 1$ as a factor.

Class 10 Chapter 8 Mathematics Solutions Selina Concise

Here let us look into some of the important concepts of Remainder and Factor Theorem.

• Remainder Theorem Definition

• Remainder Theorem Proof

• Dividing a Polynomial by a Non-Zero Polynomial

• Remainder Theorem of a Polynomial

• Euler Remainder Theorem

• Factor Theorem Definition 

• Factor Theorem Proof

• Steps to use Factor Theorem

• Polynomial Long Division

• Synthetic Division

Now let us discuss briefly each topic from the Remainder and Factor Theorem Chapter.

Remainder Theorem Definition

The remainder theorem states that when a polynomial f(x) is divided by a linear polynomial (x - a), the remainder is the same as f(a). To put it another way, if we want to test the function f(x) for a given number 'a,' we can divide it by (x - a), and the result will be f(a). It's important to note that the remainder theorem only applies when dividing a function by a linear polynomial.

Remainder Theorem Proof

The proof for the polynomial remainder theorem is derived from the Euclidean division theorem. According to these two polynomials P(x) which is the dividend and g(x) which is the divisor, asserts the existence of a quotient Q(x) and a remainder R(x) such that

P(x) = Q(x) × g(x) + R(x) and R(x) = 0 

If the divisor g(x) = x - a, where a is a constant then R(x) = 0

In both cases, R(x)  is independent of x that is R(x) is a constant. So we get 

P(x) = Q(x) × (x - a) + R

Now let us make x equal to ‘a’ in this formula, we get

P(a) = Q(a) × (a- a) + R

P(x) = Q(a) × 0 + R

P(x) = R

Hence Proved.

Dividing a Polynomial by a Non-Zero Polynomial

Here let us look at the steps to divide a polynomial by a non zero polynomial.

• To begin, arrange the polynomials in order of decreasing degree.

• To get the first term of the quotient, divide the first term of the dividend by the first term of the divisor.

• To get the remainder, multiply the divisor by the first term of the quotient and deduct this product from the dividend.

• The remainder is now the dividend, and the divisor will stay the same.

• Repeat the first step until the new dividend's degree is equal to or less than the divisor's degree.
  
  

Euler Remainder Theorem

According to Euler’s Remainder theorem states that if p and n are coprime positive integers, then

\[P \varphi (n) = 1 (mod n)\]

Where \[\varphi (n) = n(1 - \frac{1}{a}) (1 - \frac{1}{b}) . . . . . .\]

Factor Theorem Definition

The factor theorem is a theorem that connects a polynomial's factors and zeros of the polynomial. The polynomial remainder theorem is a special case of this factor theorem. Factoring a polynomial and finding the roots of a polynomial equation are two general applications of the factor theorem. The fact that these problems are practically identical is a direct consequence of the theorem.

According to factor theorem a polynomial f(x) has a factor (x-a) if and only if f(a) = 0.

Factor Theorem Proof

To prove the factor theorem we will use the same polynomial equation which we used to prove the remainder theorem.

Consider the polynomial P(x), where Q(x) is a quotient, g(x) is a divisor and R(x) is a remainder.

P(x) = Q(x) × g(x) + R(x)

If the divisor g(x) = x - a divides the polynomial P(x) such that the remainder R(x) = 0. We get

P(x) = Q(x) × (x - a) + 0

P(x) = Q(x) × (x - a)

Hence the proof.

Polynomial Long Division Method

Polynomial long division is a generalised version of the well-known arithmetic technique of long division that divides a polynomial by another polynomial of the same or lower degree. It's easy to do by hand because it breaks down a complex division problem into smaller parts.

Steps in the polynomial long division method are:

• Divide the numerator's first term by the denominator's first term.

• Multiply the denominator by the answer and add it below the numerator.

• To construct a new polynomial, subtract.
  
  

Synthetic Division

Synthetic division is a shortcut for dividing polynomials by a linear factor with a leading coefficient of 1. The process begins with decreasing the leading coefficient.

Steps in the Synthetic Division Method are:

• The synthetic division should be set up.

• Bring the leading coefficient all the way down to the bottom row.

• Multiply c by the value on the bottom row that you just wrote.

• Add the column you made in step 3 to the table.

• Repeat the process until the remainder is zero.
  
  

Importance of ICSE Class 10 Mathematics Chapter 8 Selina Concise Solutions

The ICSE Class 10 Mathematics Chapter 8 Selina concise solutions are created by Math experts working at Vedantu with only one goal in mind, that is to improve the performance of Class 10 students appearing for the Maths ICSE board exams. These study materials have been created by keeping in mind the class standards and aptitude of all the students who appear for the ICSE Class 10 examination. The ICSE Class 10 Mathematics Chapter 8 Selina Concise Solutions contains all the crucial topics from Chapter 8 like the remainder theorem, factor theorem and their applications, and much more. 

The problems are solved in a stepwise and simple way that any student can understand. The main goal of the math experts working at Vedantu is not just to help Class 10 ICSE students score good marks in the board examination, but also to make sure they learn the theories forever and do not forget anything in the long run.  All of the problems provided in the ICSE Class 10 Mathematics Chapter 8 Selina Concise solutions are solved in the right way by keeping ICSE's regulations and marking scheme in mind. This will help students obtain better marks in the final examinations.

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Conclusion

The concepts of Selina Concise Mathematics Class 10 Solutions Chapter 8 are developed to provide a comprehensive learning experience for students. These solutions to important questions are prepared by the experts after doing extensive research so that students can easily understand the concepts and clear their doubts before the exams. The experts have made sure that all the Class 10 Maths Solutions Chapter 8 Selina Concise are explained in a detailed step-by-step manner so that students will understand all the concepts clearly and ace their exams.