Question

a4 - b4+ 2b2 - 1 factorise it​

Answer 1

Your Answer:

$\tt a^{4}-b^{4}+2b^{2}-1\\\\\tt =a^{4}-(b^{4}-2b^{2}+1)\\\\\tt =a^{4}-[(b^{2})^{2}-2b^{2}(1)+(1)^{2}]\\\\\tt=(a^{2})^{2}-(b^{2}-1)^{2}\:\:\:\:\:\:\:\: [\because (a-b)^{2}=a^{2}-2ab+b^{2}]\\\\\tt = [a^{2}-(b^{2}-1)] [a^{2}+(b^{2}-1)] \:\:\:\:\:\:\:\: [ \because \: \: a^{2}-b^{2}=(a-b)(a+b)]\\\\\tt =[a^{2}-b^{2}+1] [a^{2}+b^{2}-1]$

So the factors are (a²-b²+1)(a²+b²+1)

The formulas used here are

$\tt \blacktriangleright [(a-b)^{2}=a^{2}-2ab+b^{2}]\\\\ \blacktriangleright [ a^{2}-b^{2}=(a-b)(a+b)]$

Answer 2

$\longrightarrow(a^2-b^2+1)(a^2+b^2-1)$

★ identity

$\pink\longrightarrow{\underline{\fbox{[(a-b)^2=a^2-2ab+b^2]}}}\\ \longrightarrow\pink{\underline{\fbox{ [(a^2-b^2)=(a+b)(a-b)]}}}$

★ sOLUTIOn

$\leadsto a^4-b^4+2b^2+1\\ \leadsto a^4(b^4-2b^2+1)\\ \leadsto a^4[(a^2)^2-2b^2(1)+1^2]\\ \leadsto (a^2)^2-(b^2-1)\\ \leadsto [(a^2-(b^2-1)][a^2+(b^2-1)]\\ \leadsto (a^2-b^2+1)(a^2+b^2-1)$