Chapter 6.5, Problem 21PPS

To determine

To find: the factor of the given polynomial and check whether the polynomial is factored or prime.

Answer to Problem 21PPS

  $64x^4+x{y}^3=x\left(4x+y\right)\left(16x^2+4xy+y^2\right)$ .

Explanation of Solution

Given:

The given polynomial is $64x^4-x{y}^3$ .

Calculation:

Now, to find the factor of the given polynomial. Therefore,

  $\begin{array}{l}64x^4+x{y}^3\left\{\text{given polynomial}\right\}\\ =x\left(64x^3+y^3\right)\\ =x\left({\left(4x\right)}^3+{\left(y\right)}^3\right)\left\{\text{factor out the GCF}\right\}\end{array}$

Since are both perfect cubes, so write the factor of the sum of two cubes.

  $\begin{array}{l}{\left(4x\right)}^3+{\left(y\right)}^3\\ \left\{a^3+b^3=\left(a=b\right)\left(a^2-ab+b^2\right)\right\}\\ =\left(4x+y\right)\left({\left(4x\right)}^2+\left(4x\right)y+{\left(y\right)}^2\right)\\ =\left(4x+y\right)\left(16x^2+4xy+y^2\right)\end{array}$

Thus, $64x^4+x{y}^3=x\left(4x+y\right)\left(16x^2+4xy+y^2\right)$

Hence, the factor of the given polynomial $64x^4+x{y}^3=x\left(4x+y\right)\left(16x^2+4xy+y^2\right)$ .