Exercise 14.1 Factorization- NCERT Solutions Class 8

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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

Video 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, \(x \times x \times y \times y = {x^2}{y^2}\)

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

Video 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 \(y\)

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

Video 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:

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

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

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

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

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

  
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