# Exercise 14.1 Factorization- NCERT Solutions Class 8

## Chapter 14 Ex.14.1 Question 1

Find the common factors of the terms

(i) $$12x,\;\,36$$

(ii) $$2y,\;\,22xy$$

(iii) $$14pq,\;\,28{p^2}{q^2}$$

(iv) $$2x,\;\,3{x^2},\;\,4$$

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

(vi) $$16{x^3},\; - 4{x^2},\;32x$$

(vii) $$10pq,\;20qr,\;30rp$$

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

### Solution

What is known:

Terms.

What is unknown:

Common factors of given terms.

Reasoning:

First we will find factors of each terms then find out which factors are common in each term.

Steps:

\begin{align}({\rm{i}})\quad 12 x &= 2 \times 2 \times 3 \times x\\36 &= 2 \times 2 \times 3 \times 3\end{align}

The common factors are $$2, 2, 3.$$

And, $$~2\times 2\times 3=12$$

\begin{align}({\rm{ii}})\quad 2y &= 2 \times y\\22xy &= 2 \times 11 \times x \times y\end{align}

The common factors are $$2, y.$$

And, $$2 \times y = 2y$$

\begin{align}({\rm{iii}})\quad {\rm{14 }}pq &= {\rm{2}} \times {\rm{7}} \times p \times q\\{\rm{28}}{p^{\rm{2}}}{q^{\rm{2}}} &= 2 \times 2 \times 7 \times p \times p \times q \times q \end{align}

The common factors are $$2, 7, p, q.$$

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

\begin{align}({\rm{iv}})\quad 2x &= {\rm{2}} \times x\\3x^2 &= 3 \times x \times x\\4 &= 2 \times 2\end{align}

The common factor is $$1.$$

\begin{align}({\rm{v}}) \quad 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 \\12{a^2}b &= 2 \times 2 \times 3 \times a \times a \times b \end{align}

The common factors are $$2,\, 3,\, a,\, b.$$

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

\begin{align}({\rm{vi}})\quad16{x^3} &= 2 \times 2 \times 2 \times 2 \times x \times x \times x \\ - 4{x^2} &= - 1 \times 2 \times 2 \times x \times x\\32x &= 2 \times 2 \times 2 \times 2 \times 2 \times x \end{align}

The common factors are $$2,\, 2,\, x.$$

And,$$2\times 2\times x=4x$$

\begin{align}({\rm{vii}})\quad 10pq &= 2 \times 5 \times p \times q\\20\,qr &= 2 \times 2 \times 5 \times q \times r\\30\,rp &= 2 \times 3 \times 5 \times r \times p\end{align}

The common factors are $$2, \;5.$$

And, $$2 \times 5 = 10$$

\begin{align}({\rm{viii}})\quad 3{x^2}{y^3} &= 3 \times x \times x \times y \times y \times y \\ 10{x^3}{y^2} &= 2 \times 5 \times x \times x \times x \times y \times y \\6{x^2}{y^2}z &= 2 \times 3 \times x \times x \times y \times y \times z \end{align}

The common factors are  $$x, \,x,\, y, \,y.$$

And, $$x \times x \times y \times y = {x^2}{y^2}$$

## Chapter 14 Ex.14.1 Question 2

Factorise the following expressions

(i) \begin{align} 7x - 42\end{align}

(ii) \begin{align} 6p - 12q\end{align}

(iii) \begin{align} 7{a^2} + 14a\end{align}

(iv) \begin{align} - 16z + 20{z^3}\end{align}

(v) \begin{align} 20\,{l^2}m + 30alm\end{align}

(vi) \begin{align} 5{x^2}y - 15x{y^2}\end{align}

(vii) \begin{align} 10{a^2} - 15{b^2} + 20{c^2}\end{align}

(viii) \begin{align} - 4{a^2} + 4ab - 4ca\end{align}

(ix) \begin{align} {x^2}yz + x{y^2}z + xy{z^2}\end{align}

(x) \begin{align} a{x^2}y + bx{y^2} + cxyz\end{align}

### Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of given algebraic expression.

Reasoning:

First we will find factors of each terms then find out which factors are common in each term and take out that common factor from expression.

Steps:

\begin{align}({\rm{i}})\quad 7x &= 7 \times x\\42 &= 2 \times 3 \times 7\end{align}

The common factor is $$7.$$

\begin{align} \therefore \quad & 7x - 42 \\ &= \left( {7 \times x} \right) - \left( {2 \times 3 \times 7} \right) \\ &= 7 \left( {x - 6} \right) \end{align}

\begin{align}{\rm{(ii) }}\quad 6p &= 2 \times 3 \times p\\12q &= 2 \times 2 \times 3 \times q\end{align}

The common factors are $$2$$ and $$3.$$

\begin{align}\therefore 6p - 12q &= \begin{Bmatrix} (2 \times 3 \times p) -\\ (2 \times 2 \times 3 \times q) \end{Bmatrix} \\&= 2 \times 3\,[p - (2 \times q)]\\&= 6\,(p - 2q)\end{align}

\begin{align}{\rm{(iii) }}\quad 7{a^2} &= 7 \times a \times a\\14a &= 2 \times 7 \times a\end{align}

The common factors are $$7$$ and $$a$$.

\begin{align}\therefore 7{a^2} + 14a &=\begin{Bmatrix} (7 \times a \times a) +\\ (2 \times 7 \times a) \end{Bmatrix} \\ &= 7 \times a\,[a + 2]\\ &= 7a\,(a + 2)\end{align}

\begin{align}{\rm{(iv) }}\quad 16z &= 2 \times 2 \times 2 \times 2 \times z\\20{z^3} &= 2 \times 2 \times 5 \times z \times z \times z\end{align}

The common factors are $$2,\, 2,$$ and $$z.$$

\begin{align}\therefore - 16z + 20{z^3}& =\begin{bmatrix} - (2 \times 2 \times 2 \times 2 \times z) + \\ (2 \times 2 \times 5 \times z \times z \times z) \end{bmatrix}\\ &= (2 \times 2 \times z) [ - (2 \times 2) + (5 \times z \times z) ] \\ &= 4z\left( { - 4 + 5{z^2}} \right)\end{align}

\begin{align}{\rm{(v)}}\quad 20\,{l^2}m &= 2 \times 2 \times 5 \times l \times l \times m \\30\,alm &= 2 \times 3 \times 5 \times a \times l \times m. \end{align}

The common factors are $$2,\, 5$$ $$l$$ and $$m.$$

\begin{align} \therefore 20{l^2}m + 30alm &=\begin{Bmatrix}(2 \times 2 \times 5 \times l \times l \times m ) +\2 \times 3 \times 5 \times a \times l \times m )\end{Bmatrix} \\&=\begin{Bmatrix} (2 \times 5 \times l \times m) \! \\ [(2 \!\times l) \!+\! (3 \! \times a)] \end{Bmatrix} \\ &= 10lm\,(2\,l + 3a)\end{align} \(\begin{align}\left( \text{vi} \right) \quad 5{{x}^{2}}y&=5\times x\times x\times y \\15x{{y}^{2}}&=3\times 5\times x\times y\times y \\\end{align}

The common factors are $$5, \,x,$$ and $$y.$$

\begin{align}\therefore 5{x^2}y - 15x{y^2} &= \begin{Bmatrix} ( 5 \times x \times x \times y)- \\ ( 3 \times 5 \times x \times y \times y) \end{Bmatrix} \\ & =\begin{Bmatrix} 5 \times x \times y \\ [x - (3 \times y)] \end{Bmatrix} \\ & = 5xy\,(x - 3y)\end{align}

\begin{align}{\rm{(vii)}}\quad 10{a^2} &= 2 \times 5 \times a \times a\\15{b^2} &= 3 \times 5 \times b \times b\\20{c^2} &= 2 \times 2 \times 5 \times c \times c\end{align}

The common factor is $$5.$$

\begin{align} & 10{a^2} - 15{b^2} + 20{c^2} \\ &= \begin{Bmatrix} (2 \times 5 \times a \times a) - (3 \times 5 \times b \times b) + \\ (2 \times 2 \times 5 \times c \times c) \end{Bmatrix} \\&= \begin{Bmatrix} 5[(2 \times a \times a) - (3 \times b \times b) + \\ (2 \times 2 \times c \times c)] \end{Bmatrix} \\&= 5\left( {2{a^2} - 3{b^2} + 4{c^2}} \right)\end{align}

\begin{align}(\rm{viii})\quad 4{a^2} &= 2 \times 2 \times a \times a\\4ab &= 2 \times 2 \times a \times b\\4ca &= 2 \times 2 \times c \times a\end{align}

The common factors are $$2, \,2,$$ and $$a$$.

\begin{align}\therefore & - \, 4{a^2} + 4ab - 4ca \\ &= \begin{Bmatrix} - (2 \times 2 \times a \times a) + (2 \times 2 \times a \times b) \\ - (2 \times 2 \times c \times a) \end{Bmatrix} \\&= \begin{Bmatrix} 2 \times 2 \times a \\ [ - (a) + b - c] \end{Bmatrix} \\&= 4a\,( - a + b - c)\end{align}

\begin{align}({\rm{ix}})\quad {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\end{align}

The common factors are $$x,\, y,$$ and $$z.$$

\begin{align}\therefore \quad & {x^2}yz + x{y^2}z + xy{z^2} \\ \\ & = \begin{Bmatrix} ( x \times x \times y \times z) + ( x \times y \times y \times z) \\+ (x \times y \times z \times z )\end{Bmatrix} \\&= \begin{Bmatrix} x \times y \times z \\ [x + y + z] \end{Bmatrix}\\&= xyz\,(x + y + z)\end{align}

\begin{align}{\rm{ (x)}} \,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\end{align}

The common factors are $$x$$ and $$y.$$

\begin{align} \quad & a{x^2}y + bx{y^2} + cxyz \\ \\ &=\begin{Bmatrix} ( a \times x \times x \times y) + b \times x \times y \times y )+ \\ (c \times x \times y \times z) \end{Bmatrix} \\&= (x \times y) [ (a \times x) + (b \times y) + (c \times z)] \\ &= xy\,(ax + by + cz)\end{align}

The common factors are $$x$$ and $$y$$

## Chapter 14 Ex.14.1 Question 3

Factorize:

(i) \begin{align} {x^2} + xy + 8x + 8y\end{align}

(ii) \begin{align} 15xy - 6x + 5y - 2\end{align}

(iii) \begin{align}ax + bx - ay - by\end{align}

(iv) \begin{align} 15pq + 15 + 9q + 25p\end{align}

(v) \begin{align}z - 7 + 7xy - xyz\end{align}

### Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of given algebraic expression.

Reasoning:

There are $$4$$ terms in each expression. First we will make pair of two terms from which we can take out common factors and convert the expression of $$4$$ terms into $$2$$ terms expression then take out common factors from remaining $$2$$ terms.

Steps:

\begin{align}({\rm{i}}) \quad & {x^2} + xy + 8x + 8y \\ &= \begin{Bmatrix} x \times x + x \times y \\ + 8 \times x + 8 \times y \end{Bmatrix} \\&= x(x + y) + 8(x + y)\\&= (x + y)(x + 8)\end{align}

\begin{align}{\rm{(ii)}} \quad & 15xy - 6x + 5y - 2 \\ &= \begin{Bmatrix} 3 \times 5 \times x \times y - 3 \times 2 \times x +\\ 5 \times y - 2 \end{Bmatrix} \\&= 3x(5y - 2) + 1(5y - 2)\\ &= (5y - 2)(3x + 1)\end{align}

\begin{align}{\rm{ (iii)}} \quad & ax + bx - ay - by \\ &= a \times x + b \times x - a \times y - b \times y \\ &= x(a + b) - y(a + b)\\ &= (a + b)(x - y)\end{align}

\begin{align}{\rm{ (iv)}} \quad &15pq + 15 + 9q + 25p \\ &= 15pq + 9q + 25p + 15\\ &= \begin{Bmatrix} 3 \times 5 \times p \times q + 3 \times 3 \times q + \\5 \times 5 \times p + 3 \times 5 \end{Bmatrix} \\&= 3q(5p + 3) + 5(5p + 3)\\&= (5p + 3)(3q + 5)\end{align}

\begin{align}({\rm{v}}) \quad & z - 7 + 7xy - xyz \\ &= \begin{Bmatrix} z - x \times y \times z - \\7 + 7 \times x \times y \end{Bmatrix} \\&= z(1 - xy) - 7(1 - xy)\\&= (1 - xy)(z - 7)\end{align}

Instant doubt clearing with Cuemath Advanced Math Program