Question

Factorize :
${p}^{2} + 2q - {q}^{2} + {r}^{2} - 1 - 2pr$

Answer 1

Answer:

$\boxed{(p-r+q-1)(p-r-q+1)}$

Step-by-step explanation:

$p^2+2q-q^2+r^2-1-2pr$

$\implies p^2+r^2-2pr-q^2-1+2q$

$\implies p^2+r^2-2pr-(q^2+1^2-2q)$

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We know that :

$\boxed{a^2+b^2-2ab=(a-b)^2}$

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$\implies (p-r)^2-(q-1)^2$

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We know that :

$\boxed{a^2-b^2=(a+b)(a-b)}$

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$\implies (p-r+q-1)(p-r-q+1)$

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$\boxed{\bf{(p-r+q-1)(p-r-q+1)}}$

This is the answer..............

Hope it helps you

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Answer 2

hi buddy
here's your answer looking for

${p}^{2} + 2q - {q}^{2} + {r}^{2} - 1 - 2pr \\ \\ = ( {p}^{2} + {r}^{2} - 2pr) + 2p - {q}^{2} - 1 \\ \\ = {(p - r)}^{2} - (1 + {q}^{2} - 2q) \\ \\ = {(p - r)}^{2} - {(1 - q)}^{2} \\ \\ = (p - r + 1 - q)(p - r - 1 + q)$
hope you are satisfied with my answer