Question

Factorise: 9x² -a² - 2ab -b²​

Answer 1

Before Solving:-

We observe that $9x^2=(3x)^2$, and $a^2+2ab+b^2=(a+b)^2$. So, it is a difference of two complete squares.

We have an identity $(a+b)(a-b)=a^2-b^2$, which will be used in solving this problem.

Solution:-

Given Polynomial

$=(3x)^2-(a+b)^2$

$=\{(3x)+(a+b)\}\{(3x)-(a+b)\}$

$=(3x+a+b)(3x-a-b)$

So, the factorization is $(3x+a+b)(3x-a-b)$.

More Information:-

• $ma+mb=m(a+b)$

→ Common factor.

• $a^2+2ab+b^2=(a+b)^2$
• $a^2-2ab+b^2=(a-b)^2$

→ Complete square of a binomial.

• $a^2-b^2=(a+b)(a-b)$

→ Difference of complete squares.

• $x^2+(a+b)x+ab=(x+a)(x+b)$

→ Product of binomial.

• $a^3+b^3=(a+b)(a^2-ab+b^2)$
• $a^3-b^3=(a-b)(a^2+ab+b^2)$

→ Sum or difference of complete cubes.

• $a^2+b^2+c^2-2(ab+bc+ca)=(a+b+c)^2$

→ Complete square of a trinomial.

• $a^3+b^3+b^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

→ Difference of cube and product.

• $a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)$

→ Sophie-Germain Identity.