Chapter 6.2, Problem 6MRE

To determine

To calculate: The factors of the trinomial $64+16c+c^2$.

Answer to Problem 6MRE

The trinomial $64+16c+c^2$ is factored as ${\left(c+8\right)}^2$.

Explanation of Solution

Given information:

The trinomial $64+16c+c^2$.

Formula used:

To factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

The square of sum of two numbers a and bis ${\left(a+b\right)}^2=a^2+2ab+b^2$.

Calculation:

Consider the trinomial $64+16c+c^2$.

Recall that to factor a polynomial split the middle term in such a way that sum of two numbers is the middle term and product of two numbers is same is product of first and last term.

Also the square of sum of two numbers a and bis ${\left(a+b\right)}^2=a^2+2ab+b^2$.

Therefore, the trinomial is rewritten as,

  $64+16c+c^2=c^2+2\left(8\right)c+8^2$

Write it in the factored form,

  $\begin{array}{c}64+16c+c^2=c^2+2\left(8\right)c+8^2\\ ={\left(c+8\right)}^2\end{array}$

Thus, thetrinomial $64+16c+c^2$ is factored as ${\left(c+8\right)}^2$.