Question

question 1-14
factor the expression, $49x^6 - 81y^2$.
\bigcirc $(7x^3 + 9y)(7x^3 - 9y)$
\bigcirc $(7x^3 - 9y)^2$
\bigcirc $(49x^3 - 81y)(49x^3 + 81y)$
\bigcirc $(49x^3 - 81y)^2$

Explanation:

Step1: Identify the formula

The expression \(49x^6 - 81y^2\) is a difference of squares. The formula for factoring a difference of squares is \(a^2 - b^2=(a + b)(a - b)\).

Step2: Rewrite the terms as squares

We can rewrite \(49x^6\) as \((7x^3)^2\) because \((7x^3)^2 = 7^2\times(x^3)^2=49x^6\), and \(81y^2\) as \((9y)^2\) because \((9y)^2 = 9^2\times y^2 = 81y^2\).

Step3: Apply the difference of squares formula

Using the formula \(a^2 - b^2=(a + b)(a - b)\) with \(a = 7x^3\) and \(b=9y\), we get \((7x^3+9y)(7x^3 - 9y)\).

Answer:

A. \((7x^3 + 9y)(7x^3 - 9y)\)