NCERT Class 8 Maths Factorisation

The chapter 14 begins with an introduction to Factorisation by recalling the concept of factors of natural numbers and algebraic expressions.Then the concept of Factorisation is explained in detail. Various methods of Factorisation such as method of common factors, factorisation by regrouping terms, factorisation using identities etc. are dealt separately.The next topic of discussion is division of algebraic expressions and various cases under it such as division of a monomial by another monomial,division of a polynomial by a monomial are discussed separately.The last topic of the chapter is error spotting.

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}