Question

If polynomial ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$ is divided by ${x^2} - 2x + k$ , the remainder comes out to be $x + a$ .Find k and a.

Answer 1

Hint: You have to do long division between ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$ and ${x^2} - 2x + k$ and find out the remainder and then equate it with $x + a$.And you can verify your answer by putting different values of x in the expression and verify.

Complete step-by-step answer:
According to the question, the polynomial ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$ when divided by ${x^2} - 2x + k$ leaves a remainder $x + a$.
Let another polynomial $p\left( x \right)$ be the quotient for the above division. Then the polynomial can be written as:
\[ \Rightarrow {x^4} - 6{x^3} + 16{x^2} - 25x + 10 = \left( {{x^2} - 2x + k} \right)p\left( x \right) + x + a{\text{ }}.....{\text{(1)}}\]
Here $p\left( x \right)$ will be a quadratic polynomial because the original polynomial is of 4th degree and when $p\left( x \right)$ will be multiplied by ${x^2} - 2x + k$, the result will be a 4th degree polynomial.
Further, the leading coefficient of $p\left( x \right)$ will be 1 because the leading coefficient of original polynomial is also 1.
So, based on this information, let $p\left( x \right) = {x^2} + bx + c$. Putting its value in equation (1), we’ll get:
\[ \Rightarrow {x^4} - 6{x^3} + 16{x^2} - 25x + 10 = \left( {{x^2} - 2x + k} \right)\left( {{x^2} + bx + c} \right) + x + a\]
Simplifying it further, we’ll get:
\[ \Rightarrow {x^4} - 6{x^3} + 16{x^2} - 25x + 10 = {x^4} + b{x^3} + c{x^2} - 2{x^3} - 2b{x^2} - 2cx + k{x^2} + kbx + kc + x + a\]
Separating ${x^4},{\text{ }}{x^3},{\text{ }}{x^2},{\text{ }}x$ and constant terms, we’ll get:
\[ \Rightarrow {x^4} - 6{x^3} + 16{x^2} - 25x + 10 = {x^4} + \left( {b - 2} \right){x^3} + \left( {c - 2b + k} \right){x^2} + \left( {kb - 2c + 1} \right)x + kc + a\]
Comparing coefficients of ${x^3}$ on both sides, we’ll get:
$
   \Rightarrow b - 2 = - 6 \\
   \Rightarrow b = - 4 \\
$
Comparing coefficients of ${x^2}$ on both sides, we’ll get:
$ \Rightarrow c - 2b + k = 16$
Putting $b = - 4$, we’ll get:

$
   \Rightarrow c + 8 + k = 16 \\
   \Rightarrow c + k = 8{\text{ }}.....{\text{(1)}} \\
$
Comparing coefficients of $x$ on both sides, we’ll get:
\[ \Rightarrow kb - 2c + 1 = - 25\]
Putting $b = - 4$, we’ll get:
\[\
   \Rightarrow - 4k - 2c = - 26 \\
   \Rightarrow c + 2k = 13{\text{ }}.....{\text{(2)}} \\
\]
Subtracting equation (1) from equation (2), we’ll get:
\[
   \Rightarrow c + 2k - c - k = 5 \\
   \Rightarrow k = 5 \\
\]
Putting this value in equation (1):
$
   \Rightarrow c + 5 = 8 \\
   \Rightarrow c = 3 \\
$
Now, comparing constant terms on both sides of the above equation, we’ll get:
$ \Rightarrow kc + a = 10$
Putting $c = 3$ and \[k = 5\], we’ll get:
$
   \Rightarrow 15 + a = 10 \\
   \Rightarrow a = - 5 \\
$

Thus the values of $k$ and $a$ are $5$ and $ - 5$ respectively.

Note: In such types of questions you can always verify your answer by putting different values of x in polynomials and using dividend=divisor x quotient + remainder. For example, put $x = 1$ we get ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$$
   = 1 - 6 \times 1 + 16 \times 1 - 25 \times 1 + 10 \\
   = 1 - 6 + 16 - 25 + 10 \\
   = - 4 \\
$
${x^2} - 2x + k$ $ = 1 - 2 + 5 = 4$ , $x + a$$ = 1 - 5 = - 4$ and ${x^2} - 4x + (8 - k) = 1 - 4 + 8 - 5 = 0$
So,
$
   - 4 = - 4X0 + ( - 4) \\
  LHS = RHS \\
 $
Hence verified.
If a polynomial $f\left( x \right)$ is divided by another polynomial $p\left( x \right)$ and remainder comes out as zero, then $p\left( x \right)$ is said to be a factor of $f\left( x \right)$ and this theorem is called factor theorem. Further, if $\left( {x - a} \right)$ is a factor of $f\left( x \right)$ then $x = a$ will be a root of $f\left( x \right) = 0$.