# 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}

Video Solution
Factorisation
Ex 14.2 | Question 5

## 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,

$$21 = \left( { - 7} \right) \times \left( { - 3} \right) \; \text{and} \\ \; \left( { - 7} \right) + \left( { - 3} \right) = - 10$$

\begin{align}\therefore \quad & {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 \quad & {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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