Factor completely. 16y^(6)-81x^(4) Answer:

Identify Structure: Identify the structure of the expression 16y^(6) - 81x^(4). This expression is a difference of two squares because both terms are perfect squares. 16y^(6) = (4y^(3))^2 81x^(4) = (9x^(2))^2 So, 16y^(6) - 81x^(4) can be written as (4y^(3))^2 - (9x^(2))^2.Apply Formula: Apply the difference of squares formula to factor the expression. The difference of squares formula is a^2 - b^2 = (a - b)(a + b). Here, a = 4y^(3) and b = 9x^(2). Using the formula, we get (4y^(3))^2 - (9x^(2))^2 = (4y^(3) - 9x^(2))(4y^(3) + 9x^(2)).Write Factored Form: Write down the factored form of the expression. The factored form of 16y^(6) - 81x^(4) is (4y^(3) - 9x^(2))(4y^(3) + 9x^(2)).

Home

Math Problems

Algebra 1

Factor quadratics: special cases

Full solution

Q. Factor completely.

\newline

16 y^{6}-81 x^{4}

\newline

Answer:

Identify Structure: Identify the structure of the expression $16y^{6} - 81x^{4}$ . This expression is a difference of two squares because both terms are perfect squares. $16y^{6} = (4y^{3})^2$ $81x^{4} = (9x^{2})^2$ So, $16y^{6} - 81x^{4}$ can be written as $(4y^{3})^2 - (9x^{2})^2$ .
Apply Formula: Apply the difference of squares formula to factor the expression. $\newline$ The difference of squares formula is $a^2 - b^2 = (a - b)(a + b)$ . $\newline$ Here, $a = 4y^{3}$ and $b = 9x^{2}$ . $\newline$ Using the formula, we get $(4y^{3})^2 - (9x^{2})^2 = (4y^{3} - 9x^{2})(4y^{3} + 9x^{2})$ .
Write Factored Form: Write down the factored form of the expression. $\newline$ The factored form of $16y^{6} - 81x^{4}$ is $(4y^{3} - 9x^{2})(4y^{3} + 9x^{2})$ .