NCERT Solutions for Class 8 Maths Chapter 12 Factorization

Question:(i) Factorise:

$12x+36$

Answer:

We have
$12x=2\times 2\times 3\times x$
$36=2\times 2\times 3\times 3$
So, we have $2\times 2\times 3$ common in both
Therefore,

$12x+36=$ $2\times 2\times 3(x+3)$

$12x+36=12(x+3)$

Question:(ii) Factorise : 22y-32z

Answer:

We have,
$22y=$ $2\times 11\times y$
$33z=$ $3\times 11\times z$
So, we have 11 commons in both
Therefore,

$22y-33z=11(2y-3z)$

Question:(iii) Factorise :

$(iii)14pq+35pqr$

Answer:

We have
$14pq=$ $2\times 7\times p\times q$
$35pqr=$ $5\times 7\times p\times q\times r$
So, we have

$7\times p\times q$ common in both
Therefore,

$14pq+35pqr=7pq(2+5r)$

Question:1(i) Find the common factors of the given terms.

$(i)12x,36$

Answer:

We have
$12x=2\times 2\times 3\times x$
$36=2\times 2\times 3\times 3$
So, the common factors between the two are

$2\times 2\times 3=12$

Question:1(ii) Find the common factors of the given terms

$(ii)2y,22xy$

Answer:

We have,
$2y=2\times y$
$22xy=2\times 11\times x\times y$
Therefore, the common factor between these two is 2y

Question:1(iii) Find the common factors of the given terms

$(iii)14pq,28p^2{q}^2$

Answer:

We have,
$14pq=2\times 7\times p\times q$
$28p^2{q}^2=2\times 2\times 7\times p\times p\times q\times q$
Therefore, the common factor is

$2\times 7\times p\times q=14pq$

Question:1(iv) Find the common factors of the given terms.

$(iv)2x,3x^2,4$

Answer:

We have,
$2x=2\times x$
$3x^2=3\times x\times x$
$4=2\times 2$
Therefore, the common factor between these three is 1

Question:1(v) Find the common factors of the given terms

$(v)6abc,24a{b}^2,12a^{2}b$

Answer:

We have,
$6abc=2\times 3\times a\times b\times c$
$24a{b}^2=2\times 2\times 2\times 3\times a\times b\times b$
$12a^{2}b=2\times 2\times 3\times a\times a\times b$
Therefore, the common factor is

$2\times 3\times a\times b=6ab$

Question:1(vi) Find the common factors of the given terms

$(vi)16x^3,-4x^2,32x$

Answer:

We have,
$16x^3=2\times 2\times 2\times 2\times x\times x\times x$
$4x^2=2\times 2\times x\times x$
$32x=2\times 2\times 2\times 2\times 2\times x$
Therefore, the common factor is

$2\times 2\times x=4x$

Question:1(vii) Find the common factors of the given terms

$(vii)10pq,20qr,30rp$

Answer:

We have,
$10pq=2\times 5\times p\times q$
$20qr=2\times 2\times 5\times q\times r$
$30rp=2\times 3\times 5\times r\times p$
Therefore, the common factors between these three is

$2\times 5=10$

Question:1(viii) Find the common factors of the given terms

$(viii)3x^2{y}^3,10x^3{y}^2,6x^2{y}^{2}z$

Answer:

We have,
$3x^2{y}^2$ $=3\times x\times x\times y\times y$
$10x^3{y}^2$ $=2\times 5\times x\times x\times x\times y\times y$
$6x^2{y}^{2}z$ $=2\times 3\times x\times x\times y\times y\times z$
Therefore, the common factors between these three are $x\times x\times y\times y=$ $x^2{y}^2$

Question:2(i) Factorise the following expressions

$(i)7x-42$

Answer:

We have,
$$\begin{gathered} 7x=7\times x \\ 42=7\times 2\times 3=7\times 6 \\ 7x-42=7x-7\times 6=7(x-6) \end{gathered}$$

Therefore, 7 is a common factor

Question:2(ii) Factorise the following expressions

$(ii)6p-12q$

Answer:

We have,
$6p=2\times 3\times p$
$12q=2\times 2\times 3\times q$
$\therefore$ on factorization

$6p-12q=(2\times 3\times p)-(2\times 2\times 3\times q)=(2\times 3)(p-2q)=6(p-2q)$

Question:2(iii) Factorise the following expressions

$(iii)7a^2+14a$

Answer:

We have,
$7a^2=7\times a\times a$
$14a=2\times 7\times a$
$\therefore$ $7a^2+14a=(7\times a\times a)+(2\times 7\times a)=(7\times a)(a+2)$
$=7a(a+2)$

Question:2(iv) Factorise the following expressions

$(iv)-16z+20z^3$

Answer:

We have,
$-16z=-1\times 2\times 2\times 2\times 2\times z$
$20z^3=2\times 2\times 5\times z\times z\times z$
$\therefore$ on factorization we get,
$-16z+20z^3=(-1\times 2\times 2\times 2\times 2\times z)+(2\times 2\times 5\times z\times z\times z)$
$=(2\times 2\times z)(-1\times 2\times 2+5\times z\times z)$
$=4z(-4+5z^2)$

Question:2(v) Factorise the following expressions

$20l^{2}m+30alm$

Answer:

We have,
$20l^{2}m=2\times 2\times 5\times l\times l\times m$
$30alm=2\times 3\times 5\times a\times l\times m$
$\therefore$ on factorization we get,
$20l^{2}m+30alm=(2\times 2\times 5\times l\times l\times m)+(2\times 3\times 5\times a\times l\times m)$
$=(2\times 5\times l\times m)(2\times l+3\times a)$
$=10lm(2l+3a)$

Question:2(vi) Factorise the following expressions

$5x^{2}y-15x{y}^2$

Answer:

We have,
$5x^{2}y=5\times x\times x\times y$
$15x{y}^2=3\times 5\times x\times y\times y$
$\therefore$ on factorization we get,
$5x^{2}y-15x{y}^2=(5\times x\times x\times y)-(3\times 5\times x\times y\times y)$
$=(5\times x\times y)(x-3\times y)$
$=5xy(x-3y)$

Question:2(vii) Factorise the following expressions

$10a^2-15b^2+20c^2$

Answer:

We have,
$10a^2=2\times 5\times a\times a$
$15b^2=3\times 5\times b\times b$
$20c^2=2\times 2\times 5\times c\times c$
$\therefore$ on factorization we get,
$10a^2-15b^2+20c^2=(2\times 5\times a\times a)-(3\times 5\times b\times b)+(2\times 2\times 5\times c\times c)$ $=5(2\times a\times a-3\times b\times b+2\times 2\times c\times c)$
$=5(2a^2-3b^2+4c^2)$

Question:2(viii) Factorise the following expressions

$-4a^2+4ab-4ca$

Answer:

We have,
$-4a^2=-1\times 2\times 2\times a\times a$
$4ab=2\times 2\times a\times b$
$4ca=2\times 2\times c\times a$
$\therefore$ on factorization we get,
$-4a^2+4ab-4ca=(-1\times 2\times 2\times a\times a)+(2\times 2\times a\times b)-(2\times 2\times c\times a)$

$=(2\times 2\times a)(-1\times a+b-c)$
$=4a(-a+b-c)$

Question:2(ix) Factorise the following expressions

$x^{2}yz+x{y}^{2}z+xy{z}^2$

Answer:

We have,
$x^{2}yz=x\times x\times y\times z$
$x{y}^{2}z=x\times y\times y\times z$
$xy{z}^2=x\times y\times z\times z$
Therefore, on factorization we get,
$x^{2}yz+x{y}^{2}z+xy{z}^2=(x\times x\times y\times z)+(x\times y\times y\times z)+(x\times y\times z\times z)$

$=(x\times y\times z)(x+y+z)$
$=xyz(x+y+z)$

Question:2(x) Factorise the following expressions

$a{x}^{2}y+bx{y}^2+cxyz$

Answer:

We have,
$a{x}^{2}y=a\times x\times x\times y$
$bx{y}^2=b\times x\times y\times y$
$cxyz=c\times x\times y\times z$
Therefore, on factorization we get,
$a{x}^{2}y+bx{y}^2+cxyz=(a\times x\times x\times y)+(b\times x\times y\times y)+(c\times x\times y\times z)$ $=(x\times y)(a\times x+b\times y+c\times z)$

$=xy(ax+by+cz)$

Question:3(i) Factorise $x^2+xy+8x+8y$

Answer:

We have,
$x^2=x\times x$
$xy=x\times y$
$8x=8\times x$
$8y=8\times y$
Therefore, on factorization we get,
$x^2+xy+8x+8y=(x\times x)+(x\times y)+(8\times x)+(8\times y)$
$=x(x+y)+8(x+y)$
$=(x+8)(x+y)$

Question:3(ii) Factorise

$15xy-6x+5y-2$

Answer:

We have,
$15xy=3\times 5\times x\times y$
$6x=2\times 3\times x$
$5y=5\times y$
$2=2$
Therefore, on factorization we get,
$15xy-6x+5y-2=(3\times 5\times x\times y)-(2\times 3\times x)+(5\times y)-2$
$=(5\times y)(3\times x+1)-2(3\times x+1)$
$=(5y-2)(3x+1)$

Question:3(iii) Factorise

$ax+bx-ay-by$

Answer:

We have,
$ax+bx-ay-by=a(x-y)-b(x-y)$
$=(a-b)(x-y)$
Therefore, on factorization we get,
$(a-b)(x-y)$

Question:3(iv) Factorise

$15pq+15+9q+25p$

Answer:

We have,
$15pq+15+9q+25p=5p(3q+5)+3(3q+5)$
$=(3q+5)(5p+3)$
Therefore, on factorization we get,
$(3q+5)(5p+3)$

Question:3(v) Factorise

$z-7+7xy-xyz$

Answer:

We have,
$z-7+7xy-xyz=z(1-xy)-7(1-xy)$
$=(1-xy)(z-7)$
Therefore, on factorization we get,
$(1-xy)(z-7)$

Question:1(i) Factorise the following expressions

$a^2+8a+16$

Answer:

We have,
$a^2+8a+16=a^2+4a+4a+16$
$=a(a+4)+4(a+4)$
$=(a+4)(a+4)=$ $(a+4{)}^2$
Therefore,
$a^2+8a+16=(a+4{)}^2$

Question:1(ii) Factorise the following expressions

$p^2-10p+25$

Answer:

We have,
$p^2-10p+25=p^2-5p-5p+25$
$=p(p-5)-5(p-5)$
$=(p-5)(p-5)=$ $(p-5{)}^2$
Therefore,
$p^2-10p+25=(p-5{)}^2$

Question:1(iii) Factorise the following expressions

$25m^2+30m+9$

Answer:

We have,
$25m^2+30m+9=25m^2+15m+15m+9$
$=5m(5m+3)+3(5m+3)$
$=(5m+3)(5m+3)=$ $(5m+3{)}^2$
Therefore,
$25m^2+30m+9=(5m+3{)}^2$

Question:1(iv) Factorise the following expressions

$49y^2+84yz+36z^2$

Answer:

We have,
$49y^2+84yz+36z^2$ $=49y^2+42yz+42yz+36z^2$
$=7y(7y+6z)+6z(7y+6z)$
$=(7y+6z)(7y+6z)=$ $(7y+6z{)}^2$
Therefore,
$49y^2+84yz+36z^2=(7y+6z{)}^2$

Question:1(v) Factorise the following expressions

$4x^2-8x+4$

Answer:

We have,
$4x^2-8x+4$ $=4x^2-4x-4x+4$
$=4x(x-1)-4(x-1)$
$$\begin{gathered} =4(x-1)(x-1) \\ =4(x-1{)}^2 \end{gathered}$$

Question:1(vi) Factorise the following expressions

$121b^2-88bc+16c^2$

Answer:

We have,
$121b^2-88bc+16c^2$ $=121b^2-44bc-44bc+16c^2$
$=11b(11b-4c)-4c(11b-4c)$
$=(11b-4c)(11b-4c)=$ $(11b-4c{)}^2$
Therefore,
$121b^2-88bc+16c^2$ $=$ $(11b-4c{)}^2$

Question:1(vii) Factorise the following expressions

$(l+m{)}^2-4lm$

Answer:

We have,
$(l+m{)}^2-4lm$ = $l^2+2ml+m^2-4lm$ $(using(a+b{)}^2=a^2+2ab+b^2)$
= $l^2-2lm+m^2$
= $(l-m{)}^2$ $(using(a-b{)}^2=a^{{}_2}-2ab+b^2)$

Question:2(ii) Factorise the following expressions

$63a^2-112b^2$

Answer:

We have,
$63a^2-112b^2$ $=7$ $(9a^2-16b^2)$ $=7$ $((3a{)}^2-(4b{)}^2)$ $=7(3a-4b)(3a+4b)$
$(using(a{)}^2-(b{)}^2=(a-b)(a+b))$

Question:2(iii) Factorise

$49x^2-36$

Answer:

This can be factorised as follows
$49x^2-36$ = $(7x{)}^2-(6{)}^2$ $=(7x-6)(7x+6)$ $(using(a{)}^2-(b{)}^2=(a-b)(a+b))$

Question:2(iv) Factorise

$16x^5-144x^3$

Answer:

The given question can be factorized as follows
$16x^5-144x^3$ $=16x^3(x^2-9)$
$=16x^3((x{)}^2-(3{)}^2)$ $=16x^3(x-3)(x+3)$ $(using(a{)}^2-(b{)}^2=(a-b)(a+b))$

Question:2(v) Factorise

$(l+m{)}^2-(l-m{)}^2$

Answer:

We have,
$(l+m{)}^2-(l-m{)}^2$ $=[(l+m)-(l-m)][(l+m)+(l-m)]$ (using $a^2-b^2=(a-b)(a+b)$ )
$=(l+m-l+m)(l+m+l-m)$
$=(2m)(2l)=4ml$

Question:2(vi) Factorise

$9x^2{y}^2-16$

Answer:

We have,
$9x^2{y}^2-16$ = $(3xy{)}^2-(4{)}^2$ (using $(a{)}^2-(b{)}^2=(a-b)(a+b)$ )
$=(3xy-4)(3xy+4)$

Question:2(vii) Factorise

$(x^2-2xy+y^2)-z^2$

Answer:

We have,
$(x^2-2xy+y^2)-z^2$ = $(x-y{)}^2-z^2$ $(using(a-b{)}^2=a^2-2ab+b^2)$
$=(x-y-z)(x-y+z)$ $(using(a{)}^2-(b{)}^2=(a-b)(a+b))$

Question:2(viii) Factorise

$25a^2-4b^2+28bc-49c^2$

Answer:

We have,
$25a^2-4b^2+28bc-49c^2$ = $25a^2-(2b-7c{)}^2$ $(using(a-b{)}^2=a^2-2ab+b^2)$
= $(5a{)}^2-(2b-7c{)}^2$ $(using(a{)}^2-(b{)}^2=(a-b)(a+b))$
$=(5a-(2b-7c))(5a+(2b-7c)$ )
$=(5a-2b+7c)(5a+2b-7c)$

Question:3(i) Factorise the following expressions

$a{x}^2+bx$

Answer:

We have,
$a{x}^2=a\times x\times x$
$bx=b\times x$
Therefore,
$a{x}^2+bx$ $=(a\times x\times x)+(b\times x)$
$=x(a\times x+b)$
$=x(ax+b)$

Question:3(ii) Factorise the following expressions

$7p^2+21q^2$

Answer:

We have,
$7p^2=7\times p\times p$
$21q^3=3\times 7\times q\times q$
Therefore,
$7p^2+21q^2$ $=(7\times p\times p)+(3\times 7\times q\times q)$
$=7$ $(p^2+3q^2)$

Question:3(iii) Factorise the following expressions

$2x^3+2x{y}^2+2x{z}^2$

Answer:

We have,
$2x^3=2\times x\times x\times x$
$2x{y}^2=2\times x\times y\times y$
$2x{z}^2=2\times x\times z\times z$
Therefore,
$2x^3+2x{y}^2+2x{z}^2$ $=(2\times x\times x\times x)+(2\times x\times y\times y)+(2\times x\times z\times z)$
$=(2\times x)[(x\times x)+(y\times y)+(z\times z)]$
$=2x(x^2+y^2+z^2)$

Question:3(iv) Factorise the following expressions

$a{m}^2+b{m}^2+b{n}^2+a{n}^2$

Answer:

We have,
$a{m}^2+b{m}^2+b{n}^2+a{n}^2$ $=m^2(a+b)+n^2(a+b)$
$=(a+b)$ $(m^2+n^2)$

Question:3(v) Factorise the following expressions

$(lm+l)+m+1$

Answer:

We have,
$(lm+l)+m+1$ $=lm+l+m+1$
$=l(m+1)+1(m+1)$
$=(m+1)(l+1)$

Question:3(vi) Factorise the following expressions

$y(y+z)+9(y+z)$

Answer:

We have,
$y(y+z)+9(y+z)$
Take ( y+z) common from this
Therefore,
$y(y+z)+9(y+z)$ $=(y+z)(y+9)$

Question:3(vii) Factorise the following expressions

$5y^2-20y-8z+2yz$

Answer:

We have,
$5y^2-20y-8z+2yz$ $=5y(y-4)+2z(y-4)$
$=(y-4)(5y+2z)$
Therefore,
$5y^2-20y-8z+2yz$ $=(y-4)(5y+2z)$

Question:3(viii) Factorise

$10ab+4a+5b+2$

Answer:

We have,
$10ab+4a+5b+2$ $=2a(5b+2)+1(5b+2)$
$=(5b+2)(2a+1)$
Therefore,
$10ab+4a+5b+2$ $=(5b+2)(2a+1)$

Question:3(ix) Factorise the following expressions

$6xy-4y+6-9x$

Answer:

We have,
$6xy-4y+6-9x$ $=2y(3x-2)-3(3x-2)$
$=(3x-2)(2y-3)$
Therefore,
$6xy-4y+6-9x$ $=(3x-2)(2y-3)$

Question:4(i) Factorise $a^4-b^4$

Answer:

We have,
$a^4-b^4$ = $(a^2{)}^2-(b^2{)}^2=(a^2-b^2)(a^2+b^2)=(a-b)(a+b)(a^2+b^2)$
$using(x^2-y^2)=(x-y)(x+y)$

Question:4(ii) Factorise $p^4-81$

Answer:

We have,
$p^4-81$ =
$$\begin{gathered} (p^2{)}^2-(9{)}^2=(p^2-9)(p^2+9) \\ .=(p^2-(3{)}^2)(p^2+9) \\ .=(p-3)(p+3)(p^2+9) \end{gathered}$$ $using{a}^2-b^2=(a-b)(a+b)$

Question:4(iii) Factorise $x^4-(y+z{)}^4$

Answer:

We have,
$x^4-(y+z{)}^4$ =
$$\begin{gathered} (x^2{)}^2-((y+z{)}^2{)}^2=(x^2-(y+z{)}^2)(x^2+(y+z{)}^2) \\ \Rightarrow (x-(y+z))(x+(y+z))(x^2+(y+z{)}^2) \end{gathered}$$
$(using{a}^2-b^2=(a-b)(a+b))$

Question:4(iv) Factorise $x^4-(x-z{)}^4$

Answer:

We have,
$x^4-(x-z{)}^4$ = $(x^2{)}^2-((x-z{)}^2{)}^2$ $using{a}^2-b^2=(a-b)(a+b)$
= $(x^2-(x-z{)}^2)(x^2+(x-z{)}^2)$
= $(x+(x-z))(x-(x-z))(x^2+(x-z{)}^2)$
= $(2x-z)(z)$ $($ $x^2+(x-z{)}^2$ $)$

Question:4(v) Factorise $a^4-2a^2{b}^2+b^4$

Answer:

We have,
$a^4-2a^2{b}^2+b^4$ = $a^4-a^2{b}^2-a^2{b}^2+b^4$
= $a^2(a^2-b^2)-b^2(a^2-b^2)$
= $(a^2-b^2)(a^2-b^2)$ $using{a}^2-b^2=(a-b)(a+b)$
= $(a^2-b^2{)}^2$
= $((a-b)(a+b){)}^2$
= $(a-b{)}^2(a+b{)}^2$