Question

nomial function by factoring completely. (don they come up!) 12. $f(x) = -6x^3 + 26x^2 + 20x$ (with handwritten steps and solutions including factoring and finding roots: $-2x(3x^2 - 13x - 10)$, then factoring further to $-2x(3x + 2)(x - 5) = 0$ and roots $x = 0, -\frac{2}{3}, 5$)

Explanation:

Step1: Factor out the GCF

Given the polynomial \( f(x)=-6x^{3}+26x^{2}+20x \), the greatest common factor (GCF) of the terms is \(-2x\). Factoring out \(-2x\) gives:
\( f(x)=-2x(3x^{2}-13x - 10) \)

Step2: Factor the quadratic

Now, factor the quadratic \( 3x^{2}-13x - 10 \). We need two numbers that multiply to \( 3\times(-10)=-30 \) and add up to \(-13\). The numbers are \(-15\) and \(2\). Rewrite the middle term:
\( 3x^{2}-15x + 2x-10 \)
Group the terms:
\( (3x^{2}-15x)+(2x - 10) \)
Factor out the GCF from each group:
\( 3x(x - 5)+2(x - 5) \)
Now, factor out \((x - 5)\):
\( (3x + 2)(x - 5) \)

Step3: Write the fully factored form

Substitute the factored quadratic back into the expression:
\( f(x)=-2x(3x + 2)(x - 5) \)

Step4: Find the roots

To find the roots, set \( f(x) = 0 \):
\( -2x(3x + 2)(x - 5)=0 \)
Using the zero - product property (if \( ab = 0 \), then either \( a = 0 \), \( b = 0 \) or \( c = 0 \) for \( a\times b\times c=0 \)):

• If \(-2x=0\), then \( x = 0 \).
• If \( 3x+2 = 0 \), then \( 3x=-2\) and \( x=-\frac{2}{3}\).
• If \( x - 5=0\), then \( x = 5 \).

Answer:

\( x = 0 \), \( x = -\frac{2}{3} \), \( x = 5 \)