(𝑝^2−2𝑝𝑞+𝑞^2 )−𝑟^2 We know that, (𝑎−𝑏)^2=𝑎^2−2𝑎𝑏+𝑏^2 𝑎^2−𝑏^2=(𝑎+𝑏)(𝑎−𝑏) nbsp; ⇒(𝑝^2−2𝑝𝑞+𝑞^2 )−𝑟^2 =(𝑝−𝑞)^2−𝑟^2 =(𝑝−𝑞−𝑟)(𝑝−𝑞+𝑟) ...

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Factorisation by Common Factors

Factorise: 𝑝...

Question

Factorise: $(p^{2} - 2 p q + q^{2}) - r^{2}$

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Solution

$(p^{2} - 2 p q + q^{2}) - r^{2}$

We know that,
$(a - b)^{2} = a^{2} - 2 a b + b^{2} a^{2} - b^{2} = (a + b) (a - b)$

$\Rightarrow (p^{2} - 2 p q + q^{2}) - r^{2} = (p - q)^{2} - r^{2} = (p - q - r) (p - q + r)$

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