# Ex.14.1 Q2 Factorization - NCERT Maths Class 8

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## Question

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}

Video Solution
Factorisation
Ex 14.1 | Question 2

## Text 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 $$y$$

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