Question

for exercises 23–38, find the zeros of the function and state the multiplicities. (see examples 2–4 on pp. 189–192) 26. ( k(x) = -6x^3 + 26x^2 - 28x )

Explanation:

Step1: Factor out the GCF

First, factor out the greatest common factor (GCF) from the polynomial \( k(x)=-6x^{3}+26x^{2}-28x \). The GCF of \(-6x^{3}\), \(26x^{2}\), and \(-28x\) is \(-2x\). So we have:
\( k(x)=-2x(3x^{2}-13x + 14) \)

Step2: Factor the quadratic

Next, factor the quadratic \(3x^{2}-13x + 14\). We need two numbers that multiply to \(3\times14 = 42\) and add up to \(-13\). The numbers are \(-6\) and \(-7\). Rewrite the middle term:
\(3x^{2}-6x-7x + 14\)
Group the terms:
\((3x^{2}-6x)+(-7x + 14)=3x(x - 2)-7(x - 2)=(3x - 7)(x - 2)\)
So now, \(k(x)=-2x(3x - 7)(x - 2)\)

Step3: Find the zeros

To find the zeros, set each factor equal to zero:

• For \(-2x = 0\), we get \(x = 0\)
• For \(x - 2 = 0\), we get \(x = 2\)
• For \(3x - 7 = 0\), we get \(x=\frac{7}{3}\)

Step4: Determine multiplicities

Since each factor is linear (degree 1), the multiplicity of each zero is 1.

Answer:

The zeros of the function \(k(x)\) are \(x = 0\) (multiplicity 1), \(x = 2\) (multiplicity 1), and \(x=\frac{7}{3}\) (multiplicity 1).