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

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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