Question

use the algebra tiles to determine the polynomial equivalent to this expression. \\(\frac{x^2 + 6x + 9}{x + 3}\\) \\(\bigcirc\\) a. \\(x^2 + 3x\\) \\(\bigcirc\\) b. \\(x - 3\\) \\(\bigcirc\\) c. \\(x + 3\\) \\(\bigcirc\\) d. \\(x^2 - 3x\\)

Explanation:

Step1: Factor the numerator

The numerator is \(x^{2}+6x + 9\). We know that \(x^{2}+6x + 9=(x + 3)^{2}\) (since \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(b = 3\), \(2ab=6x\), \(a^{2}=x^{2}\), \(b^{2}=9\)).

Step2: Simplify the fraction

Now we have the expression \(\frac{(x + 3)^{2}}{x + 3}\). For \(x
eq - 3\) (to avoid division by zero), we can cancel out one factor of \(x + 3\) from the numerator and the denominator. So \(\frac{(x + 3)^{2}}{x + 3}=x + 3\).

Answer:

C. \(x + 3\)