Chapter 11.3, Problem 5STP

What are the factors of the expression $x^2+12x-64$?

What number would you change to make it a perfect square? Explain.

To determine

To find the factors and to make it perfect square of the expression

Answer to Problem 5STP

The factors are $\left(x+16\right)\left(x-4\right)$ and 36 number is to make the perfect square expression

Explanation of Solution

Given information:

The expression is $x^2+12x-64$

Calculation:

The factors of the expression can be obtained as

  $\begin{array}{l}{x}^2+12x-64\\ =x^2+\left(16x-4x\right)-64\left[\text{write 12}x=\left(16x-4x\right)\right]\\ =x^2+16x-4x-64\\ =x\left(x+16\right)-4\left(x+16\right)\left[\begin{array}{l}\text{the first and second term has a common factor,}x\text{ and}\\ \text{the third and fourth term has a common factor,4}\end{array}\right]\\ =\left(x+16\right)\left(x-4\right)\end{array}$

The factor are $\left(x+16\right)\left(x-4\right)$

To make it perfect square, we change the constant term $-64$

  $\begin{array}{l}{x}^2+12x-64\\ =x^2+2\left(6\right)\left(x\right)+{\left(6\right)}^2\left[\text{perfect square trinomial}\right]\\ =x^2+12x+36\\ ={\left(x+6\right)}^2\end{array}$

Hence,

The factors are $\left(x+16\right)\left(x-4\right)$ and 36 number is to make the perfect square expression