# The area of a rectangle painting is given by the trinomial a^2-14a+45. The painting length is...

## Question:

The area of a rectangle painting is given by the trinomial {eq}a^2-14a+45 {/eq}. The painting length is (a+2). What is the painting width?

## Painting Dimensions:

The rectangle's area can be expressed with its length and width. These dimensions are flexible and can have constant unknown variables. In solving these unknown variables for the possible values, we should always consider the nature of the given. Since we are dealing with a painting per se, dimensions should have positive values only.

The area of a rectangle painting is given by the trinomial {eq}a^2-14a+45 {/eq}. The painting length is (a+2).

The area of the rectangle is given by:

\begin{align} Area&=Length \times Width && \left [ \text{Substitute the given conditions} \right ]\\[0.2cm] a^2-14a+45&=\left ( a+2 \right ) \times Width && \left [ \text{Solving for the width} \right ]\\[0.2cm] \displaystyle Width&=\frac{a^2-14a+45}{\left ( a+2 \right )}\\[0.2cm] \displaystyle Width&=\frac{\left ( a-5 \right )\left ( a-9 \right )}{\left ( a+2 \right )} && \left [ a\neq-2 \ \text{and} \ a> 9 \right ]\\[0.2cm] \end{align}

Therefore, the {eq}\displaystyle Width=\frac{\left ( a-5 \right )\left ( a-9 \right )}{\left ( a+2 \right )} {/eq} wherein {eq}a\neq-2 \ \text{and} \ a> 9 {/eq} since we cannot divide a fraction with 0 and that a should be greater than 9 to provide positive values for the dimensions.