Question

Factorize: $\frac{4}{9}{p}^2–1.$

• A. (2/3p-1)(2/3-1)
• B. (2/3p+1)(2/3p-1)
• C. (4/9p+1)(4/9p-1)
• D. (4/9p-1)(4/9p-1)

Answer 1

The correct option is

A

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

The difference between the squares of any two terms = (The sum of the two terms) $\times$ (Their difference)

$a^2–b^2=(a+b)(a-b)$

The algebraic expression is$\frac{4}{9}{p}^2–1.$

$=(\frac{2}{3}p{)}^2-(1{)}^2$

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