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) \(\begin{align}  {{(l + m)}^2} - 4lm \text{ (Hint: Expand }{{\left( {l + m} \right)}^2}\;{\rm{ first}}) \end{align}\)

(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:

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

\[\begin{align} \left[ \text {Using identity}\;{(x + y)^2} = {x^2} + 2xy + {y^2},\\\! {\text{considering }} x = a\;{\rm{ and}} \;y = {\rm{4}} \right] \end{align}\]

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

\[ \left[ \text {Using identity}\;(a - b)^2 = {a^2} - 2ab + {b^2},\\\!\text{considering } a = p \; \rm{and} \;y = {\rm{5}} \right]\]

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

\[\left[\text {Using identity}\;{(a + b)^2} = {a^2} + 2ab + {b^2},\\\!{\text{considering }} a = 5m\;{\rm{ and}}\; b = {\rm{3}}\right]\]

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

\[\left[ \text{Using identity }\;{(a + b)^2} = {a^2} + 2ab + {b^2},\\\!{\rm{considering}}\;a = 7\;y\;{\rm{and}}\;b = 6 \right]\]

\(\begin{align}(\rm{v}) \quad 4{x^2} - 8x + 4 &= {{(2x)}^2} - 2(2x)(2) + {{(2)}^2}\\&= {{(2x - 2)}^2} \\&= {{[(2)(x - 1)]}^2} = 4{{(x - 1)}^2}\end{align}\)
\[\left[ \text{Using identity }{(a - b)^2} = {a^2} - 2ab + {b^2},\\\!\text{Considering }\;a = 2x\;{\rm{and}}\;y = 2\right]\]

\(\begin{align}(\rm{vi})\quad 121{b^2} - 88bc + 16{c^2} &= {{(11b)}^2} - 2(11b)(4c) + {{(4c)}^2}\\
&= {(11b - 4c)}^2\end{align}\)
\[\left[\text{Using identity } \;{(a - b)^2} = {a^2} - 2ab + {b^2},\\\!{\text{Considering }}\;a = 11b\;{\rm{and}}\;b = 4c\right]\]

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

\[\left[\text{Using identity}\;{(a - b)^2} = {a^2} - 2ab + {b^2},\\\! {\text{considering}}\;a = l\;{\rm{and}}\;b = m\right]\]

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

\[\left[\text{Using identity }\;(x + y)^2 = {x^2} + 2xy + {y^2},\\\!\text{considering }\;x = {a^2}\;{\rm{and}}\;y = {b^2}\right]\]

  
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