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Important Questions for CBSE Class 8 Maths Chapter 14 - Factorisation

Last updated date: 29th May 2024
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CBSE Class 8 Maths Important Questions for Factorisation - Free PDF Download

The solutions for Important Questions For Class 8 Maths Chapter 14 factorization chapter are prepared by the experts according to NCERT curriculum. These Important questions on Factorization chapter important questions include finding factors, factorizing the given equations, division of factors. Class 8 Maths Chapter 14 Extra Questions are helpful for the students to streamline their knowledge by practising the extra questions provided on Factorization.

Free PDF download of Important Questions with solutions for CBSE Class 8 Chapter 14 - Factorisation prepared by expert Mathematics teachers from the latest edition of CBSE(NCERT) books.

Vedantu is a platform that provides free (CBSE) NCERT Solution and other study materials for students. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 8 Science , Maths solutions and solutions of other subjects.

Every student has a different perception of maths. While some find maths interesting, there are many others who start disliking the subject as they keep getting promoted in higher classes. No wonder, a lot of students of Class 8 also have this problem; and this can be solved only by developing a clear concept of Maths with the help of a proper guide. To understand factorization, students need to solve as many factorisation Class 8 questions as possible from both the textbook as well as the solutions available on any credible online platform like Vedantu. Through solving various short and long factorisation questions Class 8, the student will gain confidence in factorization.

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Access Important Questions For Class 8 Maths Chapter 14 – Factorisation

Very Short Answer Questions (1 Mark)

1. Factorise $\text{12}{{\text{a}}^{\text{2}}}\text{b+15a}{{\text{b}}^{\text{2}}}$

(a) $\text{3ab(4a+5b)}$

(b) $\text{3ab}$

(c) $\text{4a+5b}$

(d) $\text{3ab(5a+4b)}$

Ans: $\text{12}{{\text{a}}^{\text{2}}}\text{b = 2}\times \text{2}\times \text{3}\times \text{a}\times \text{a}\times \text{b}$

$\text{15a}{{\text{b}}^{\text{2}}}\text{ = 3}\times \text{5}\times \text{a}\times \text{b}\times \text{b}$

Taking common factor from each term,

$12{{a}^{2}}b+15a{{b}^{2}}=3ab(4a+5b)$

Hence the correct option is (a).

2. When we factorise an expression, we write it as a ______ of factors.

Ans: When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

3. Factorise 6xy – 4y + 6 – 9x

(a) (3x – 2)

(b) (3x – 2)(2y – 3)

(c) (2y – 3)

(d) (2x – 3)(3y – 2)

Ans: $6xy-4y+6-9x=2\times 3\times x\times y-2\times 2\times y+2\times 3-3\times 3\times x$

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

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

Hence the correct option is (b).

4. Find the common factors of 12x, 36

(a) 12

(b) 36

(c) x

(d) 12x

Ans: Here, $\text{12x = 2}\times \text{2}\times \text{3}\times x$

$\text{36 = 2}\times \text{2}\times \text{3}\times 3$

So, common factor =$\text{2}\times \text{2}\times \text{3}$

$\Rightarrow 4\times 3=12$

Hence the correct option is (a).

5. Solve $\text{7}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{{\text{z}}^{\text{2}}}\text{ }\!\!\div\!\!\text{ 14xyz}$

(a) 2

(b) 4x

(c) 3y

(d) $\dfrac{\text{xyz}}{\text{2}}$

Ans: $\text{7}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{{\text{z}}^{\text{2}}}\text{ }\!\!\div\!\!\text{ 14xyz = 7}\times \text{x}\times \text{x}\times \text{y}\times \text{y}\times \text{z}\times \text{z}\div 2\times 7\times x\times y\times z$

$\text{= x}\times \text{y}\times \text{z}\div 2$

$\text{=}\dfrac{\text{xyz}}{\text{2}}$

Hence the correct option is (d).

6. Factorise $\text{49}{{\text{p}}^{\text{2}}}\text{-36}$

(a) (7p+6)(7p+6)

(b) (7p-6)(7p-6)

(c) (7p-6)(7p+6)

(d) (6p-7)(7p-6)

Ans: Using the identity,${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$

$\text{49}{{\text{p}}^{\text{2}}}\text{-36 = (7p}{{\text{)}}^{2}}\text{ - }{{\text{6}}^{2}}$

$\text{= (7p+6)(7p-6)}$

Hence the correct option is (c).

7. Factorise $\text{a}{{\text{x}}^{\text{2}}}\text{+b}{{\text{y}}^{\text{2}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+a}{{\text{y}}^{\text{2}}}$

Ans: $\text{a}{{\text{x}}^{\text{2}}}\text{+b}{{\text{y}}^{\text{2}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+a}{{\text{y}}^{\text{2}}}=\text{a}{{\text{x}}^{\text{2}}}\text{+b}{{\text{x}}^{\text{2}}}\text{+a}{{\text{y}}^{\text{2}}}\text{+b}{{\text{y}}^{\text{2}}}$

$\text{= a}\times \text{x}\times \text{x+b}\times \text{x}\times \text{x+a}\times \text{y}\times \text{y+b}\times \text{y}\times \text{y}$

$\text{= }{{\text{x}}^{2}}(a+b)+{{y}^{2}}(a+b)$

$\text{= }(a+b)({{x}^{2}}+{{y}^{2}})$

8. Factorise using identity ${{\text{x}}^{\text{2}}}\text{+10x+25}$

Ans: ${{\text{x}}^{\text{2}}}\text{+10x+25 = }{{\text{x}}^{\text{2}}}\text{+2}\times \text{5}\times \text{x+}{{\text{5}}^{2}}$

Using the identity, ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

$\text{= }{{\text{x}}^{\text{2}}}\text{+2}\times \text{5}\times \text{x+}{{\text{5}}^{2}}$

$\text{= (x+5}{{\text{)}}^{2}}$

$\text{= (x+5)(x+5)}$

9. Factorise ${{\text{(a+b)}}^{\text{2}}}\text{- (a-b}{{\text{)}}^{\text{2}}}$

Ans:${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

${{\text{(a+b)}}^{\text{2}}}\text{- (a-b}{{\text{)}}^{\text{2}}}=\text{ (}{{\text{a}}^{2}}\text{+2ab+}{{\text{b}}^{2}}\text{) - (}{{\text{a}}^{2}}\text{-2ab+}{{\text{b}}^{2}}\text{)}$

$={{\text{a}}^{2}}\text{+2ab+}{{\text{b}}^{2}}\text{ - }{{\text{a}}^{2}}\text{+2ab-}{{\text{b}}^{2}}$

$=\text{4ab}$

10. Simplify $\left[ \text{1}{{\text{0}}^{\text{2}}}\text{-18 }\!\!\times\!\!\text{ 10+81} \right]$

Ans:$\left[ \text{1}{{\text{0}}^{\text{2}}}\text{-18 }\!\!\times\!\!\text{ 10+81} \right]=\left[ {{10}^{2}}-2\times 9\times 10+{{9}^{2}} \right]$

Using the identity, ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

$=\left[ {{10}^{2}}-2\times 9\times 10+{{9}^{2}} \right]$

$={{\left[ 10-9 \right]}^{2}}$

$={{\left[ 1 \right]}^{2}}$

$=1$

11. Factorise $\text{16}{{\text{a}}^{\text{2}}}\text{-}\dfrac{\text{25}}{\text{4}{{\text{a}}^{\text{2}}}}$

Ans:$16{{a}^{2}}-\dfrac{25}{4{{a}^{2}}}={{(4a)}^{2}}-{{\left( \dfrac{5}{2a} \right)}^{2}}$

Using the identity,${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$

$={{(4a)}^{2}}-{{\left( \dfrac{5}{2a} \right)}^{2}}$

$=\left( 4a+\dfrac{5}{2a} \right)\left( 4a-\dfrac{5}{2a} \right)$

12. Simplify $\dfrac{\text{4}{{\text{m}}^{\text{2}}}\text{-169}{{\text{n}}^{\text{2}}}}{\text{2m+13n}}$

Ans:$\dfrac{\text{4}{{\text{m}}^{\text{2}}}\text{-169}{{\text{n}}^{\text{2}}}}{\text{2m+13n}}\text{=}\dfrac{{{\text{(2m)}}^{\text{2}}}\text{-(13n}{{\text{)}}^{\text{2}}}}{\text{(2m+13n)}}$

Using the identity,${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$

$\text{=}\dfrac{{{\text{(2m)}}^{\text{2}}}\text{- (13n}{{\text{)}}^{\text{2}}}}{\text{(2m+13n)}}$

$\text{=}\dfrac{\text{(2m-13n) - (2m+13n)}}{\text{(2m+13n)}}$

$\text{=(2m-13n)}$

13. Simplify $\dfrac{\left( {{\text{p}}^{\text{2}}}\text{+11p+28} \right)}{\left( \text{p+4} \right)}$

Ans: Here, the numerator can be further factorised as follow

$\text{(}{{\text{p}}^{\text{2}}}\text{+11p+28) = (}{{\text{p}}^{\text{2}}}\text{+7p+4p+28)}$

$=p(p+7)+4(p+7)$

$=(p+7)(p+4)$

Now,$\dfrac{\left( {{\text{p}}^{\text{2}}}\text{+11p+28} \right)}{\left( \text{p+4} \right)}=\dfrac{(p+7)(p+4)}{(p+4)}$

$=(p+7)$

14. Find x if $\text{x(y-z)=}\dfrac{\text{(4}{{\text{y}}^{\text{2}}}\text{-4}{{\text{z}}^{\text{2}}}\text{)}}{\text{(y+z)}}$

Ans: Taking 4 common from the numerator we get

$\text{x(y-z)=}\dfrac{\text{4(}{{\text{y}}^{\text{2}}}\text{- }{{\text{z}}^{\text{2}}}\text{)}}{\text{(y+z)}}$

Using the identity,${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$ in the numerator

$\text{x(y-z)=}\dfrac{\text{4(y-z)(y+z)}}{\text{(y+z)}}$

$\text{x(y-z) = 4(y-z)}$

On comparing both side we can see that x=4.

15. Find x and y that xy = 72, x + y = 17.

Ans: Here we have been given xy and x + y

To find out the value of x and y we can the identity ${{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$

${{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$

${{(17)}^{2}}={{x}^{2}}+2(72)+{{y}^{2}}$

${{(17)}^{2}}-144={{x}^{2}}+{{y}^{2}}$

${{x}^{2}}+{{y}^{2}}=289-144$

${{x}^{2}}+{{y}^{2}}=145$

Now, putting the values in identity ${{(x-y)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$

${{(x-y)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$

${{(x-y)}^{2}}=145-2(72)$

${{(x-y)}^{2}}=145-144$

$x-y=1$

Adding both the equations i.e., x + y = 17 and x – y = 1

(x + y)+(x - y) = 17+1

x + y + x – y = 18

2x = 18

x = 9

Putting the value of x in equation x + y = 17

x + y = 17

9 + y = 17

y = 17 – 9

y = 8

Hence the value of x is 9 and y is 8.

16. Find the remainder in the following $\text{(}{{\text{x}}^{\text{4}}}\text{-}{{\text{a}}^{\text{4}}}\text{) }\!\!\div\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{)}$.

Ans: For dividing both the equation first we have to simplify it.

So, $({{x}^{4}}-{{a}^{4}})$ can be written as ${{({{x}^{2}})}^{2}}-{{({{a}^{2}})}^{2}}$

Now by applying the identity ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$

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

Now, $\dfrac{\left( {{\text{x}}^{\text{4}}}\text{-}{{\text{a}}^{\text{4}}} \right)}{\text{(}{{\text{x}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{)}}\text{=}\dfrac{\text{(}{{\text{x}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{)(}{{\text{x}}^{\text{2}}}\text{-}{{\text{a}}^{\text{2}}}\text{)}}{\text{(}{{\text{x}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{)}}$

$\dfrac{\left( {{\text{x}}^{\text{4}}}\text{-}{{\text{a}}^{\text{4}}} \right)}{\text{(}{{\text{x}}^{\text{2}}}\text{+}{{\text{a}}^{\text{2}}}\text{)}}\text{=(}{{\text{x}}^{\text{2}}}\text{-}{{\text{a}}^{\text{2}}}\text{)}$

Hence the remainder is $\text{(}{{\text{x}}^{\text{2}}}\text{-}{{\text{a}}^{\text{2}}}\text{)}$.

17. Simplify $\dfrac{{{\text{(0}\text{.87)}}^{\text{2}}}\text{-(0}\text{.13}{{\text{)}}^{\text{2}}}}{\text{(0}\text{.87-0}\text{.13)}}$

Ans: By applying the identity ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ in the numerator.

$\dfrac{{{\text{(0}\text{.87)}}^{\text{2}}}\text{-(0}\text{.13}{{\text{)}}^{\text{2}}}}{\text{(0}\text{.87-0}\text{.13)}}=\dfrac{\text{(0}\text{.87+0}\text{.13)-(0}\text{.87-0}\text{.13}{{\text{)}}^{\text{2}}}}{\text{(0}\text{.87-0}\text{.13)}}$

$\dfrac{{{\text{(0}\text{.87)}}^{\text{2}}}\text{-(0}\text{.13}{{\text{)}}^{\text{2}}}}{\text{(0}\text{.87-0}\text{.13)}}=\text{(0}\text{.87+0}\text{.13)}$

$\dfrac{{{\text{(0}\text{.87)}}^{\text{2}}}\text{-(0}\text{.13}{{\text{)}}^{\text{2}}}}{\text{(0}\text{.87-0}\text{.13)}}=1$

18. Using identity ${{\text{(a-b)}}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{-2ab+}{{\text{b}}^{\text{2}}}$ find the value of ${{\text{(98)}}^{\text{2}}}$ .

Ans: ${{\text{(98)}}^{\text{2}}}$ can be further written as ${{\text{(100 - 2)}}^{\text{2}}}$

By comparing the above equation with the identity, we get

a = 100 and b = 2

${{\text{(98)}}^{\text{2}}}={{(100)}^{2}}-2(100)(2)+{{(2)}^{2}}$

$=10000-400+4$

$=9604$

19. Using identity ${{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{=(a+b)(a-b)}$ . Find ${{(1.02)}^{2}}-{{(0.98)}^{2}}$

Ans: Using identity ${{\text{a}}^{\text{2}}}\text{-}{{\text{b}}^{\text{2}}}\text{=(a+b)(a-b)}$

${{\text{(1}\text{.02)}}^{\text{2}}}\text{-(0}\text{.98}{{\text{)}}^{\text{2}}}=(1.02+0.98)(1.02-0.98)$

$=(2)(0.04)$

$=(0.08)$

20. Simplify $\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-16)}}{\text{(}{{\text{a}}^{\text{2}}}\text{-25)}}\text{ }\!\!\div\!\!\text{ }\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-2a-8)}}{\text{(}{{\text{a}}^{\text{2}}}\text{+10a+25)}}$

Ans: Identities to be used in the question are;

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

${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ $\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-16)}}{\text{(}{{\text{a}}^{\text{2}}}\text{-25)}}\text{ }\!\!\div\!\!\text{ }\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-2a-8)}}{\text{(}{{\text{a}}^{\text{2}}}\text{+10a+25)}}=\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-}{{\text{4}}^{2}}\text{)}}{\text{(}{{\text{a}}^{\text{2}}}\text{-}{{\text{5}}^{2}}\text{)}}\text{ }\!\!\div\!\!\text{ }\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-2a-8)}}{\text{(}{{\text{a}}^{\text{2}}}\text{+2}\times \text{5}\times \text{a+}{{\text{5}}^{2}}\text{)}}$

$=\dfrac{\text{(a+4)(a-4)}}{\text{(a+5)(a-5)}}\text{ }\!\!\div\!\!\text{ }\dfrac{\text{(}{{\text{a}}^{\text{2}}}\text{-4a+2a-8)}}{\text{(}{{\text{a}}^{\text{2}}}\text{+2}\times \text{5}\times \text{a+}{{\text{5}}^{2}}\text{)}}$

$=\dfrac{\text{(a+4)(a-4)}}{\text{(a+5)(a-5)}}\text{ }\!\!\div\!\!\text{ }\dfrac{\text{a(a-4)+2(a-4)}}{{{\text{(a+5)}}^{2}}}$

$=\dfrac{\text{(a+4)(a-4)}}{\text{(a+5)(a-5)}}\times \dfrac{\text{(a+5)(a+5)}}{\text{(a+2)(a-4)}}$

$=\dfrac{\text{(a+5)(a+4)}}{\text{(a-5)(a+2)}}$

21. Simplify ${{\text{(xy+yz)}}^{\text{2}}}\text{-(xy-yz}{{\text{)}}^{\text{2}}}$

Ans: Using the identities ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

$\text{=(xy}{{\text{)}}^{\text{2}}}\text{+(yz}{{\text{)}}^{\text{2}}}\text{+2(xy)(yz) - }\left[ {{(xy)}^{2}}+{{(yz)}^{2}}-2(xy)(yz) \right]$

$\text{=(xy}{{\text{)}}^{\text{2}}}\text{+(yz}{{\text{)}}^{\text{2}}}\text{+2(xy)(yz) - }{{(xy)}^{2}}-{{(yz)}^{2}}+2(xy)(yz)$

$\text{=2(x}{{\text{y}}^{2}}\text{z)+}2(x{{y}^{2}}z)$

$\text{=4x}{{\text{y}}^{2}}\text{z}$

22. The area of a rectangle is $6{{a}^{2}}+36a$ and its width is 36a. Find its length.

Ans: Let the length of the rectangle is x

Breath = 36a

$\text{Area of rectangle = Length }\!\!\times\!\!\text{ Breath}$

$\text{6}{{\text{a}}^{\text{2}}}\text{+36a = x}\times \text{36a}$

$\dfrac{\text{6}{{\text{a}}^{\text{2}}}\text{+36a}}{36a}\text{ = x}$

$x=\dfrac{\text{6a(a+6)}}{36a}\text{ }$

$x=\dfrac{\text{(a+6)}}{6}\text{ }$

Hence the length of rectangle is $\dfrac{\text{(a+6)}}{6}\text{ }$

23. Match the following:

 a. $\text{25}{{\text{p}}^{\text{2}}}\text{-}{{\text{q}}^{\text{2}}}$ 1.  ${{\text{x}}^{\text{2}}}\text{-2x-35}$ b. ${{\left( \text{x-1} \right)}^{\text{2}}}\text{-36}$ 2. $\left( \text{5p-q} \right)\left( \text{5p+q} \right)$ c. $\text{dj+hj+dq+hq}$ 3. $\dfrac{{{\text{e}}^{\text{5}}}}{{{\text{m}}^{\text{2}}}{{\text{r}}^{\text{5}}}}\left( \dfrac{\text{1}}{{{\text{m}}^{\text{2}}}}\text{+}\dfrac{{{\text{e}}^{\text{4}}}}{{{\text{r}}^{\text{4}}}} \right)$ d. $\dfrac{{{\text{e}}^{\text{5}}}}{{{\text{m}}^{\text{4}}}{{\text{r}}^{\text{5}}}}\text{+}\dfrac{{{\text{e}}^{\text{9}}}}{{{\text{m}}^{\text{2}}}{{\text{r}}^{\text{9}}}}$ 4. $\left ( d+h \right )\left ( q+j \right )$ e. $\dfrac{\text{12}{{\text{p}}^{\text{2}}}\text{+132p}}{\text{12p}}$ 5. $\left( \text{p+11} \right)$

Ans:

1. Using the identity: ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$

$\text{25}{{\text{p}}^{\text{2}}}\text{-}{{\text{q}}^{\text{2}}}\text{ = }{{\left( \text{5p} \right)}^{2}}\text{-}{{\text{q}}^{2}}$

$=\left( \text{5p-q} \right)\left( \text{5p+q} \right)$  ⇒ Option (2).

1. ${{\left( \text{x-1} \right)}^{\text{2}}}\text{-36 = }\left( {{\text{x}}^{2}}-2x+1 \right)-36$

$=\text{ }{{x}^{2}}-2x-35$ $\Rightarrow$ Option (1).

1. $\text{dj+hj+dq+hq = d}\left( j+q \right)+h(j+q)$

$=\left( \text{d+h} \right)\left( \text{q+j} \right)$ $\Rightarrow$ Option (4).

1. $\dfrac{{{\text{e}}^{\text{5}}}}{{{\text{m}}^{\text{4}}}{{\text{r}}^{\text{5}}}}\text{+}\dfrac{{{\text{e}}^{\text{9}}}}{{{\text{m}}^{\text{2}}}{{\text{r}}^{\text{9}}}}\text{ = }\dfrac{{{\text{e}}^{\text{5}}}}{{{\text{m}}^{2}}{{\text{r}}^{\text{5}}}}\left( \dfrac{\text{1}}{{{\text{m}}^{\text{2}}}}\text{+}\dfrac{{{\text{e}}^{\text{4}}}}{{{\text{r}}^{\text{4}}}} \right)$ $\Rightarrow$ Option (3).

2. $\dfrac{\text{12}{{\text{p}}^{\text{2}}}\text{+132p}}{\text{12p}}\text{ = }\dfrac{12p\left( p+11 \right)}{\text{12p}}$ $\Rightarrow$ Option (5).

24. The combined area of two squares is $\text{20c}{{\text{m}}^{\text{2}}}$. Each side of one square is twice as long as a side of the other square. Find the length of the sides of each square.

Ans:

Let the side of the smaller square be S and that of the bigger square be 2S.

Combined area of the two squares = $\text{20c}{{\text{m}}^{\text{2}}}$

${{S}^{2}}+{{(2S)}^{2}}=20$

${{S}^{2}}+4{{S}^{2}}=20$

$5{{S}^{2}}=20$

${{S}^{2}}=4$

$S=2cm$

Hence the side of the smaller square is 2cm and that of the bigger square is 4cm.

25. Find the factors of $\text{25}{{\text{x}}^{\text{2}}}\text{-4}{{\text{y}}^{\text{2}}}\text{+28yz-49}{{\text{z}}^{\text{2}}}$

Ans: $\text{25}{{\text{x}}^{\text{2}}}\text{-4}{{\text{y}}^{\text{2}}}\text{+28yz-49}{{\text{z}}^{\text{2}}}=\text{25}{{\text{x}}^{\text{2}}}\text{-(4}{{\text{y}}^{\text{2}}}\text{-28yz+49}{{\text{z}}^{\text{2}}})$

$=\text{25}{{\text{x}}^{\text{2}}}\text{- }\!\![\!\!\text{ (2y}{{\text{)}}^{\text{2}}}\text{-2}\times 2\text{y}\times \text{7z+(7z}{{\text{)}}^{\text{2}}}]$

Using the identity ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

${{(5x)}^{2}}-{{(2y-7z)}^{2}}$

Now using the identity  ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$

$(5x+2y-7z)(5x-2y+7y)$

Maths Chapter 14 - Factorization

Factorization consists of writing a number or mathematical expression as a product of several factors, usually smaller or simpler expressions of the same kind.

Ex: We can write the equation 4x + 4 in a simpler form by using factorization as 4(x + 1) which is the same as the original expression but in a simplified form.

Factorization can be done on natural numbers and algebraic expressions.

Factorization on Natural numbers

Factorizing a natural number is easy as compared to algebraic expressions. Most natural numbers factorization involves prime factors.

Ex: 45 can be written in a factorized form as 15*3.

65 can be written as 13*5.

Factorization on Algebraic Expressions

Algebraic expression involves linear equations, quadratic equations, cubic equations and higher-order polynomials. Factorization of these algebraic expressions involves varieties of methods. Simple linear equations can be factorized easily, but cubic equations and higher-order polynomials require standard identities and formulas to factorize them.

Ex: 3xy + 3y can be factorized as 3y(x + 1).

X2 + y2 + 2xy can be factorized as (x + y)2 using standard identities.

Factorization methods

• Method of common factors

• Factorisation by regrouping terms

• Factorisation using identities

Method of Common Factors

If we find the variables of two or more numbers, and then find that some variables are "common", then those factors are called common factors.

Ex:

1. Factorize 5a2b2 + 15ab

Ans: By using the Method of common factors, we can simplify the equation as follow:

5a2b2 can be written as 5*a*a*b*b

15ab can be written as 5*3*a*b

From these two terms, 5*a*b is the common factor.

So 5*a*b (a*b +3) are the factors of the expression 5a2b2 + 15ab.

2. Factorize 4x4 + 8x3 + 12x2 + 20x

Ans: By using the Method of common factors, we can simplify the equation as follow:

4x4 can be written as 4*x*x*x*x

8x3 can be written as 4*2*x*x*x

12x2 can be written as 4*3*x*x

20x can be written as 4*5*x

From these four terms in the expression 4*x is the common factor.

So the given equation can be factorized as follows:

4*x (x3 + 2x2 + 3x+ 5) are the factors of the expression 4x4 + 8x3 + 12x2 + 20x.

Factorisation by Regrouping Terms

• The terms of the given expression must be arranged in appropriate groups in factorization by regrouping, in such a way that all groups have a common factor.

• Factor out each group.

• Take out the variable that is common to each group.

Ex:

1. Factorize 6xy – 4y + 6 – 9x

Ans: First check whether the given expression has common factors if not then go for regrouping method. Here we group the first 2 terms and last 2 terms.

Regrouping first 2 terms

In 6xy - 4y, 2y can be taken as the common term

So 2y (3x -2) is the first term.

Now regrouping last 2 terms

Here 3 is the common factor in 6-9x

So 3 (2-3x) is the second term which can be simplified to -3 (3x-2)

Now putting first and second term together

2y (3x -2) - 3 (3x-2)

Here 3x-2 is common in both the terms,

So (3x-2) (2y-3) are the factors of the expression 6xy – 4y + 6 – 9x.

Factorisation Using Identities

Factorization by using identities will help us to find the factors of an algebraic expression in simple steps.

The common identities used are as follows:

• (a + b)2 = a2 + 2ab +b2

• (a - b)2 = a2 - 2ab +b2

• a2 – b2 = (a + b)(a – b)

Ex:

1. Factorize 4x2 - 16x + 16 by using identities.

Ans: If we observe the given expression it is of the form a2 - 2ab +b2. So let us compare the given expression with the identity to find the terms a and b.

By comparing 4x2 - 16x + 16 with a2 - 2ab +b2

a2 = 4x so a = 2x

b2 = 16 so b = 4

2ab = 16x

The factors a2 - 2ab +b2 is (a - b)2 by identity.

So the factors of 4x2 - 16x + 16 is (2x - 4)2.

2. Factorize 100 - 49x2 using identities.

Ans: If we carefully the given expression is of the form a2 – b2. So let us compare the given expression with identity to find the terms a and b.

By comparing 100 - 49x2 with a2 – b2

a2 = 100 so a = 10

b2 = 49x2 so b = 7x

The factor of a2 – b2 is (a + b)(a – b).

So the factors of 100 - 49x2 is (10 + 7x) (10 - 7x).

Division of Algebraic Expression

• Division of a monomial by a monomial

• Division of a polynomial by a monomial

• Division of a polynomial by a polynomial

Division of a Monomial By a Monomial

In algebra, a monomial is an expression containing one term. Monomials contain numbers that are multiplied together, whole numbers and variables.

Ex: 2x, 3x2, 7x4y2

Example problems:

1. Divide 10x3 by 2x2.

Ans: 10x3 can be written as 2*5*x*x*x.

2x2 can be written as 2*x*x.

So 10x3 / 2x2 = 2*5*x*x*x / 2*x*x = 5x.

2. Divide 49x4y3z2 by 7x3y2z.

Ans: 49x4y3z2 can be written as 7*7*x*x*x*x*y*y*y*z*z

7x3y2z can be written as 7*x*x*x*y*y*z.

So 49x4y3z2 / 7x3y2z = 7*7*x*x*x*x*y*y*y*z*z / 7*x*x*x*y*y*z = 7xyz.

Division of a Polynomial by a Monomial

A polynomial is an equation made up of variables and coefficients, which contains only the addition, subtraction, multiplication, and non-negative integer exponentiation operations of variables.

Ex: 4x2 - 16x + 16, (10 + 7x) (10 - 7x), 4x4 + 8x3 + 12x2 + 20x.

Example problems:

1. Divide the polynomial 4a3 + 16a2 + 32a by a monomial 4a.

Ans: Here we have to divide each term of the polynomial in the numerator by the monomial in the denominator.

4a3 can be written as 4*a*a*a. Dividing 4a3 by 4a, we get

4*a*a*a / 4*a = a2…………………(1)

16a2 can be written as 4*4*a*a. Dividing 16a2 by 4a, we get

4*4*a*a / 4*a = 4a…………………(2)

32a can be written as 4*8*a. Dividing 32a by 4a, we get

4*8*a / 4*a = 8……………………..(3)

Now adding equations (1), (2) and (3) we get the final answer as follows:

a2 + 4a + 8

Division of a Polynomial by a Polynomial

Here the polynomial in the numerator is divided by the different/ same polynomial in the denominator.

Example problem:

1.  Divide the polynomial 7x2 + 14x by another polynomial x + 2.

Ans: First let us factorize the numerator and simplify and divide by denominator equation.

7x2 can be written as 7*x*x.

14x can be written as 7*2*x.

So 7*x is common factor in the numerator, so

7x2 + 14x = 7x (x + 2)

Now dividing 7x (x + 2) by the denominator x + 2. x + 2 is common in both numerator and denominator so it will get cancelled. Therefore the division of 7x2 + 14x by x + 2 is

7x (x + 2) / (x + 2) = 7x.

What are the Benefits of Important Questions from Vedantu for Class 8 Maths Chapter 14 - Factorisation

• Focus on key topics for efficient studying.

• Prepares students for exams and reduces anxiety.

• Reinforces understanding of fundamental concepts.

• Teaches effective time management.

• Enables self-assessment and progress tracking.

• Strategic approach for higher scores.

• Covers a wide range of topics for comprehensive understanding.

• Supports exam preparation and boosts confidence.

Conclusion

Students can refer to the free PDF available at Vedantu platform to prepare for Important Questions For Class 8 Maths Chapter 14 board exams which are prepared according to NCERT curriculum. This PDF also provides extra questions as practice problems for students so that they can improve their subject knowledge on Factorization chapter and its related topics.

FAQs on Important Questions for CBSE Class 8 Maths Chapter 14 - Factorisation

1. What do you mean by factors in Maths?

In mathematics, a factor is a number or algebraic expression that evenly divides another number or expression, leaving no residual or remainder. The numbers you multiply to produce another number are called "factors." There are several factorizations for certain integers (more than one way of being factored). Factors are important components as they tell you about the property of different numbers whether they are even or odd, perfect squares and cubes or not and so on.

2. What is Factorisation?

Factorization is the process of dividing a large number into smaller numbers that when multiplied together yield the original value. Factorization is the process of dividing a number into its components or divisors. These components or factors might be integers, variables, or algebraic expressions themselves. Practice all problems related to factorization to completely grasp the fundamentals of factorization and to do well in all of your tests with the help of Vedantu.

3. What do you mean by algebraic expressions?

At least one variable and at least one operation are included in algebraic expressions (addition, subtraction, multiplication, division). To describe an algebraic expression, we use its related terms and operations on the terms to explain it. A term might be a single variable (or) a single constant (or) a multiplication or division operation that combines variables and constants. This definition is used to determine the terms in an algebraic expression.

4. What do you mean by polynomials?

Variables (or indeterminate), terms, exponents, and constants make up a polynomial expression. Polynomials may be divided into three categories. Monomial, Binomial, and Trinomial.

There is just one term in a monomial expression.

A two-term expression is known as a binomial expression.

A trinomial is an expression with three terms.

A polynomial can have any number of terms, but it cannot have an infinite number of terms. Polynomial division and other related sums are important in Chapter 14 of Class 8 Maths.

5. How can I avail the Solutions of Class 8 math Chapter 14?

The solutions are easily available on the Vedantu site.

• Visit the page NCERT Solutions for Class 8 Maths Chapter 14.

• The webpage with Vedantu’s solutions for Class 8 Maths Chapter 14 will open.