Chapter 5, Problem 11PT

To determine

To calculate: The factors of the equation, $8y^2+x^{3}y$ .

Answer to Problem 11PT

The factorized form of the equation is: $\left(2y+x\right)\left(4y^2-2yx+x^2\right)$ .

Explanation of Solution

Given information:

Calculate the factors of the equation $8y^2+x^{3}y$:

If the polynomial is not factorable then write prime.

Formula used:

The identity,

  $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$ .

Calculation:

Consider the equation,

  $8y^2+x^{3}y$

Recall that for the equation the identity is:

  $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$

Therefore, the solution of the equation is given by,

  $\begin{array}{c}8y^2+x^{3}y=\left(2y+x\right)\left({\left(2y\right)}^2-2yx+{\left(x\right)}^2\right)\\ =\left(2y+x\right)\left(4y^2-2yx+x^2\right)\end{array}$

The factorized form is: $\left(2y+x\right)\left(4y^2-2yx+x^2\right)$ here no further simplification is left.

So, This equation is factorized equation.

Thus, the factorized form is: $\left(2y+x\right)\left(4y^2-2yx+x^2\right)$ .