Question

Factorise the following:

${(2x+\frac{1}{3})}^2-{(x-\frac{1}{2})}^2$

• A. $(x-\frac{1}{6})(x-\frac{5}{6})$
• B. $(3x-\frac{1}{6})(x+\frac{5}{6})$
• C. $(3x-\frac{1}{6})(x-\frac{5}{6})$
• D. $(x-\frac{1}{6})(x+\frac{5}{6})$

Answer 1

Answer: (B)

$(3x-\frac{1}{6})(x+\frac{5}{6})$

Given, ${(2x+\frac{1}{3})}^2-{(x-\frac{1}{2})}^2$

Using $a^2-b^2=(a+b)(a-b)$

$=(2x+\frac{1}{3}+x-\frac{1}{2})(2x+\frac{1}{3}-x+\frac{1}{2})$

$=(3x+\frac{2-3}{6})(x+\frac{2+3}{6})$

$=(3x-\frac{1}{6})(x+\frac{5}{6})$