Question

Factorise
${a}^{2} - 1 - 2x - {x}^{2}$



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Answer 1

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✺Answer

♦️To factorise:

• $\large{ \rm{ {a}^{2} - 1 - 2x - {x}^{2}}}$

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✺Explanation of Q.

We know for doing factorisation, we must know the basic identities i.e algebraic identities. An identity is a expression which satisfy for all value of variable.

♠️Some basic identities:

$\large{ \rm{(a + b) {}^{2} = {a}^{2} + 2ab + {b}^{2}}} \\ \\ \large{ \rm{(a - b) {}^{2} = {a}^{2} - 2ab + {b}^{2} }} \\ \\ \large{ \rm{ {a}^{2} - {b}^{2} = (a + b)(a - b)}} \\ \\ \large{ \rm{( a + b) {}^{3} = {a}^{3} + b {}^{3} + 3ab(a + b)}} \\ \\ \large{ \rm{(a - b) {}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b)}}$

While solving this question, we will use (a + b)^2 and a^2-b^2.

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✺Solution:

$\large{ \rm{ \rightarrow {a}^{2} - 1 - 2x - {x}^{2}}} \\ \\ \rm{ \green{this \: can \: be \: written \: as....}}\\ \\ \large{ \rm{ \rightarrow {a}^{2} - (1 + 2x + {x}^{2})}} \\ \\ \rm{ \green{by \: using \: (a + b) {}^{2} = {a}^{2} + 2ab + {b}^{2} }} \\ \\ \large{ \rm{ \rightarrow {a}^{2} - \{{(x)}^{2} + 2 \times x \times 1 + (1) {}^{2} \}}} \\ \\ \large{ \rm{ \rightarrow {a}^{2} - (x + 1) {}^{2} }} \\ \\ \rm{ \green{by \: using \: {a}^{2} - {b}^{2} = (a + b)(a - b)}} \\ \\ \large{ \rm{ \rightarrow \{a + (x + 1)\}\{(a - (x + 1)\}}} \\ \\ \large{ \rightarrow \boxed{ \rm{ \red{ (a + x + 1)(a- x - 1)}}}}$

♠️Hence, Factorised!

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Answer 2

Answer:

$(a+x+1)(a-x+1)$

we arrange the terms and get

$a^{2}-(x^{2}+2x+1)$

we can write that as

$a^{2}-(x(x+1)+1(x+1))$

$a^{2}-(x+1)^{2}$

by using the formula $a^2-b^2=(a+b)(a-b)$

$a^2-(x+1)^2=(a+x+1)(a-x+1)$