Question

6 joy multiplied (3x - 3) by a second binomial. if the product was 9x² - 18x + 9, find the second binomial. a. (3x - 3) b. (3x + 3) c. (3x - 6) d. none of the above

Explanation:

Step1: Let the second binomial be \(ax + b\). We know that \((3x - 3)(ax + b)=9x^{2}-18x + 9\). First, expand the left - hand side:

Using the distributive property (FOIL method), \((3x-3)(ax + b)=3x(ax + b)-3(ax + b)=3ax^{2}+3bx-3ax - 3b=3ax^{2}+(3b - 3a)x-3b\)

Step2: Compare the coefficients of the expanded form and the given product \(9x^{2}-18x + 9\)

• For the coefficient of \(x^{2}\):

We have \(3a = 9\). Solving for \(a\), divide both sides by 3: \(a=\frac{9}{3}=3\)

• For the constant term:

We have \(-3b = 9\). Solving for \(b\), divide both sides by \(- 3\): \(b=\frac{9}{-3}=-3\)

• Let's check the coefficient of \(x\) with \(a = 3\) and \(b=-3\):

The coefficient of \(x\) in the expanded form is \(3b-3a\). Substitute \(a = 3\) and \(b=-3\): \(3\times(-3)-3\times3=-9 - 9=-18\), which matches the coefficient of \(x\) in \(9x^{2}-18x + 9\)
So the second binomial is \(ax + b=3x-3\)

Answer:

A. \((3x - 3)\)