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

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 \(z.\)

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

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

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

\[\begin{align} \therefore & 20{l^2}m + 30alm \\ &= \begin{Bmatrix} \begin{pmatrix} 2 \times 2 \times 5 \\ \times l \times l \times m \end{pmatrix} \\ + \begin{pmatrix} 2 \times 3 \times 5 \times \\ a \times l \times m \end{pmatrix} \end{Bmatrix} \\&=\begin{Bmatrix} (2 \times 5 \times l \times m) \! \\ [(2 \!\times l) \!+\! (3 \! \times a)] \end{Bmatrix} \\ &= 10lm\,(2\,l + 3a)\end{align}\]

\(\begin{align}\left( \text{vi} \right) \quad 5{{x}^{2}}y&=5\times x\times x\times y \\15x{{y}^{2}}&=3\times 5\times x\times y\times y \\\end{align}\)

The common factors are \(5, \,x,\) and \(y.\)

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

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

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 \(z.\)

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

\(\begin{align}{\rm{ (x)}} \,a{x^2}y &= a \times x \times x \times y\\bx{y^2} &= b \times x \times y \times y\\cxyz &= c \times x \times y \times z\end{align}\)

The common factors are \(x\) and \(y.\)

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

The common factors are \(x\) and \(y\)

  
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