# Ex.14.2 Q1 Factorization - NCERT Maths Class 8

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## Question

Factorize the following expressions.

(i) \begin{align} {a^2} + 8a + 16\end{align}

(ii) \begin{align} {p^2} - 10p + 25\end{align}

(iii) \begin{align} 25{m^2} + 30m + 9\end{align}

(iv) \begin{align} 49{y^2} + 84yz + 36{z^2}\end{align}

(v) \begin{align} 4{x^2} - 8x + 4\end{align}

(vi) \begin{align} 121{b^2} - 88bc + 16{c^2}\end{align}

(vii) $${{(l + m)}^2} - 4lm$$

(Hint: Expand $${{\left( {l + m} \right)}^2}$$ first)

(viii) \begin{align} {a^4} + 2{a^2}{b^2} + {b^4} \end{align}

Video Solution
Factorisation
Ex 14.2 | Question 1

## Text Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of the algebraic expression.

Reasoning: Use identity:

\begin{align} & {{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \\ & {{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}} \\ \end{align}

Steps:

\begin{align}({\rm{i}}) \,{a^2} + 8a + 16 &= \!\!{{(a)}^2} + 2 \times \!a \times 4 +\! {{(4)}^2} \\&= {{(a + 4)}^2} \end{align}

Using identity\begin{align}{(x + y)^2} = {x^2} + 2xy + {y^2}\end{align},considering $$x = a$$  and $$y = 4$$

\begin{align}({\rm{ii}}) \, & {p^2} - 10p + 25 \\ &= {(p)}^2 - 2 \times p \times 5 + {(5)}^2 \\&= {(p - 5)}^2 \end{align}

Using identity\begin{align}(a - b)^2 = {a^2} - 2ab + {b^2}\end{align},considering $$a = p$$  and $$y = 5$$

\begin{align}({\rm{iii}} ) \, & 25{m^2} + 30m + 9 \\&= {{(5m)}^2} + 2 \times 5m \times 3 + {{(3)}^2} \\&= {{(5m + 3)}^2} \end{align}

Using identity\begin{align}(a +b)^2 = {a^2} + 2ab + {b^2}\end{align},considering $$a = 5m$$  and  $$b = 3$$

\begin{align}({\rm{iv}}) \, & 49{y^2} + 84yz +36{z^2}\\&= {{(7y)}^2} + 2 \times (7y) \times (6z) + {{(6z)}^2} \\ &= {{(7y + 6z)}^2}\end{align}

Using identity\begin{align}(a +b)^2 = {a^2} + 2ab + {b^2}\end{align},considering $$a = 7y$$  and $$b = 6$$

\begin{align}({\rm{v}}) \, & 4{x^2} - 8x + 4 \\ &= {{(2x)}^2} - 2(2x)(2) + {{(2)}^2} \\&= {{(2x - 2)}^2} \\&= {{[(2)(x - 1)]}^2} \\ &= 4{{(x - 1)}^2}\end{align}

Using identity\begin{align}(a -b)^2 = {a^2} - 2ab + {b^2}\end{align},considering $$a = 2x$$  and $$y = 2$$

\begin{align}({\rm{vi}})\, &121{b^2} - 88bc + 16{c^2} \\ &={{(11b)}^2} - 2(11b)(4c) + {{(4c)}^2} \\ &= {(11b - 4c)}^2\end{align}

Using identity\begin{align}(a -b)^2 = {a^2} - 2ab + {b^2}\end{align},considering $$a = 11b$$  and $$b = 4c$$

\begin{align}({\rm{vii}})\,& {{(l + m)}^2} - 4lm \\ &= {l^2} + 2lm + {m^2} - 4lm \\&= {l^2} - 2lm + {m^2}\\&= {(l - m)}^2 \end{align}

Using identity\begin{align}(a -b)^2 = {a^2} - 2ab + {b^2}\end{align},considering $$a = l$$ and $$b = m$$

\begin{align}{\rm{(viii) }} \, & {a^4} + 2{a^2}{b^2} + {b^4} \\ &= {{\left( {{a^2}} \right)}^2} + 2\left( {{a^2}} \right)\left( {{b^2}} \right) + {{\left( {{b^2}} \right)}^2} \\&= {{\left( {{a^2} + {b^2}} \right)}^2} \end{align}

Using identity\begin{align}(x + y)^2 = {x^2} + 2xy + {y^2}\end{align},considering $$x = {a^2}$$ and $$y = {b^2}$$

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