Ex.14.2 Q5 Factorization - NCERT Maths Class 8

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Question

Factorise the following expressions

(i) \(\begin{align}{p^2} + 6p + 8\end{align}\)

(ii) \(\begin{align}{q^2} - 10q + 21\end{align}\)

(iii) \(\begin{align}{p^2} + 6p - 16\end{align}\)

Text Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of the algebraic expression.

Reasoning:

In general, for factorising an algebraic expression of the type \({{x}^{2}}+px+q\) , we find two factors a and b of q (i.e., the constant term) such that \(ab = q\) and \(a + b = p.\)

Steps:

(i) \(\quad {p^2} + 6p + 8\)

It can be observed that, \(8 = 4 \times 2\) and \(4 + 2 = 6\)

\[\begin{align}\therefore {p^2} + 6p + 8 &= {p^2} + 2p + 4p + 8\\&= p(p + 2) + 4(p + 2)\\&= (p + 2)(p + 4)\end{align}\]

(ii)\(\quad {q^2} - 10q + 21\)

It can be observed that, \(21 = \left( { - 7} \right) \times \left( { - 3} \right)\) and \(\left( { - 7} \right) + \left( { - 3} \right) = - 10\)

\[\begin{align}\therefore {q^2} - 10q + 21 &= {q^2} - 7q - 3q + 21\\ &= q(q - 7) - 3(q - 7)\\&= (q - 7)(q - 3)\end{align}\]

(iii) \(\quad {p^2} + 6p - 16\)

It can be observed that, \(16{\rm{ }} = {\rm{ }}\left( { - 2} \right){\rm{ }} \times {\rm{ }}8\) and \(8{\rm{ }} + {\rm{ }}\left( { - 2} \right){\rm{ }} = {\rm{ }}6\)

\[\begin{align}\therefore {p^2} + 6p - 16 &= {p^2} + 8p - 2p - 16\\&= p(p + 8) - 2(p + 8)\\&= (p + 8)(p - 2)\end{align}\]

  
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