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How do you factor $16{x^4} - 81{y^4}$?

seo-qna
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Answer
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Hint: In this question we need to find the factor of algebraic expression $16{x^4} - 81{y^4}$. Given algebraic expression is of two variables $x$ and $y$. To solve this question we need to use the following basic algebraic identities such as ${a^2} - {b^2} = (a + b)(a - b)$. To solve this question we need to know the square root of a number or how to find the square root of a number. To solve this we also need to know the laws of exponents.

Complete step by step solution:
Let us try to solve this question in which we are asked to find the factor of the given algebraic expression $16{x^4} - 81{y^4}$. To find the factors of the equation we manipulate the given algebraic expression by using our knowledge of exponents, so that we can apply the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$. So, let’s come back to the question.
We have to find factor of $16{x^4} - 81{y^4}$, this can be written as
$16{x^4} - 81{y^4} = {(4{x^2})^2} - {(9{y^2})^2}$ $(1)$
Because we know that from law of exponents ${a^{b \cdot c}} = {({a^b})^c}$ and also we know that $16 = {4^2}$ and $81 = {9^2}$
Now, applying the identity ${a^2} - {b^2} = (a + b)(a - b)$ in equation $(1)$, we get
${(4{x^2})^2} - {(9{x^2})^2} = (4{x^2} - 9{x^2})(4{x^2} + 9{y^2})$ $(2)$
Now, again applying the identity ${a^2} - {b^2} = (a + b)(a - b)$ in equation (2), we get
$(4{x^2} - 9{x^2})(4{x^2} + 9{y^2}) = (2x - 3y)(2x + 3y)(4{x^2} + 9{y^2})$ $(3)$
Equation $(3)$ cannot be further factorized because this equation has no more linear factors.
Hence the factor of algebraic expression $16{x^4} - 81{y^4} = (2x - 3y)(2x + 3y)(4{x^2} + 9{y^2})$.

Note: For solving this type of question in which we are asked to find the factor of algebraic expression having the knowledge of some basic algebraic identities are must such as ${a^2} - {b^2} = (a + b)(a - b)$,
${(a + b)^2} = {a^2} + 2ab + {b^2}$ etc.
To solve these types of questions we just have to break the expression using knowledge of exponents and apply known algebraic identities.