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

\( \begin{align} {\rm{(i)}}\quad &a{x^2} + bx \\ & = a \times x \times x + b \times x \\ & = x(ax + b) \end{align} \)

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

\( \begin{align} \left( {{\rm{iii}}} \right) \quad & 2{x^3} + 2x{y^2} + 2x{z^2} \\ & = 2x\left( {{x^2} + {y^2} + {z^2}} \right) \end{align} \)

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

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

\( \begin{align} \left( {{\rm{vi}}} \right)\quad &y{\rm{(}}y + z{\rm{)}} + {\rm{9(}}y + z{\rm{)}} \\ & = {\rm{(}}y + z{\rm{)(}}y + {\rm{9)}} \end{align} \)

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

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

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

  
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