Chapter 8.5, Problem 43P

The area of a rectangular painting is given by the trinomial $a^2-6a-16.$

The painting’s length is $(a+2).$

What is the painting’s width?

To determine

To find : the painting’s width.

Answer to Problem 43P

  $a-8$

Explanation of Solution

Given information: The area of a rectangular painting is given by the trinomial $a^2-6a-16$. The painting’s length is $\left(a+2\right)$ .

Calculation:

We will factorize the polynomial for the area of the rectangular painting by splitting the midle term into two terms whose sum is $-6a$ and product is $-16a^2$ .

Therefore, by splitting the middle term , we get

  $\begin{array}{l}{a}^2-6a-16\\ =a^2-8a+2a-16\\ =\left(a^2-8a\right)+\left(2a-16\right)\\ =a\left(a-8\right)+2\left(a-8\right)\\ =\left(a-8\right)\left(a+2\right)\end{array}$

We know that area of a rectangle is product of its length and width.

Therefore, width of the painting is the ratio of its area to its length.

Width= $=\frac{a^2-6a-16}{a+2}=\frac{\left(a-8\right)\left(a+2\right)}{a+2}=a-8$

Thus, width of the painting is $a-8$