# Ex.14.2 Q3 Factorization - NCERT Maths Class 8

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## Question

Factorise the expressions

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

(ii) \begin{align}7{p^2} + 21{q^2}\end{align}

(iii) \begin{align}2{x^3} + 2x{y^2} + 2x{z^2}\end{align}

(iv) \begin{align}a{m^2} + b{m^2} + b{n^2} + a{n^2}\end{align}

(v) \begin{align}(lm + l) + m + 1\end{align}

(vi) \begin{align}y(y + z) + 9(y + z)\end{align}

(vii) \begin{align}5{y^2} - 20y - 8z + 2yz\end{align}

(viii) \begin{align}10ab + 4a + 5b + 2\end{align}

(ix) \begin{align}6xy - 4y + 6 - 9x\end{align}

## Text Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of given algebraic expression.

Reasoning: For part ($$\rm{i}$$), ($$\rm{ii}$$), ($$\rm{iii}$$) and ($$\rm{vi}$$) - 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.

For part ($$\rm iv$$), ($$\rm v$$), ($$\rm vii$$), ($$\rm viii$$), ($$ix$$) - 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:

$${\rm{(i)}}\quad a{x^2} + bx = a \times x \times x + b \times x = x(ax + b)$$

$$\left( {{\rm{ii}}} \right)\quad 7{p^2} + 21{q^2} = 7 \times p \times p + 3 \times 7 \times q \times q = 7\left( {{p^2} + 3{q^2}} \right)$$

$$\left( {{\rm{iii}}} \right) \quad 2{x^3} + 2x{y^2} + 2x{z^2} = 2x\left( {{x^2} + {y^2} + {z^2}} \right)$$

\begin{align}({\rm{iv}})\quad a{m^2} + b{m^2} + b{n^2} + a{n^2} &= a{m^2} + b{m^2} + a{n^2} + b{n^2}\\&= {m^2}(a + b) + {n^2}(a + b)\\&= (a + b)\left( {{m^2} + {n^2}} \right) \end{align}

\begin{align}({\rm{v}})\quad \left( {lm + l} \right) + m + 1 &= lm + m + l + 1\\&= m\left( {l + 1} \right) + 1\left( {l + 1} \right)\\&= \left( {l + 1} \right)\left( {m + 1} \right)\end{align}

$$\left( {{\rm{vi}}} \right)\quad {\rm{ }}y{\rm{(}}y + z{\rm{)}} + {\rm{9(}}y + z{\rm{)}} = {\rm{(}}y + z{\rm{)(}}y + {\rm{9)}}$$

\begin{align}{\rm{(vii)}}\quad 5{y^2} - 20y - 8z + 2yz &= 5{y^2} - 20y + 2yz - 8z\\&= 5y(y - 4) + 2z(y - 4)\\&= (y - 4)(5y + 2z)\end{align}

\begin{align}{\rm{(viii)}}\quad10ab + 4a + 5b + 2 &= 10ab + 5b + 4a + 2\\&= 5b(2a + 1) + 2(2a + 1)\\&= (2a + 1)(5b + 2)\end{align}

\begin{align}({\rm{ix)}}\quad 6xy - 4y + 6 - 9x &= 6xy - 9x - 4y + 6\\&= 3x(2y - 3) - 2(2y - 3)\\ &= (2y - 3)(3x - 2)\end{align}

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