Question

polynomial function by factoring (complex)
n they come up!)

1. $f(x) = \frac{-6x^3}{-2x} + \frac{26x^2}{-2x} + \frac{20x}{-2x}$

$-2x(3x^2 - 13x - 10)$
$3x^2 - 15x + 2x - 10$
$\frac{3x^2}{3x} \frac{-15x}{3x} \frac{+2x}{2} \frac{-10}{2}$
$3x(x - 5) + 2(x - 5)$
$-2x(3x + 2)(x - 5) = 0$
$x = 0, -\frac{2}{3}, 5$
(side work: $a \cdot c = -30$, factors -15 and 2, sum -13 (b))

Explanation:

Step1: Factor out the greatest common factor (GCF)

First, we identify the GCF of the terms \(-6x^{3}\), \(26x^{2}\) and \(20x\). The GCF of the coefficients \(- 6\), \(26\) and \(20\) is \(- 2\) and the GCF of the variable terms \(x^{3}\), \(x^{2}\) and \(x\) is \(x\). So the GCF is \(-2x\)

We factor out \(-2x\) from the polynomial:

\(f(x)=-2x(3x^{2}-13x - 10)\)

Step2: Factor the quadratic trinomial

We need to factor the quadratic \(3x^{2}-13x - 10\). For a quadratic \(ax^{2}+bx + c\) (here \(a = 3\), \(b=- 13\) and \(c=-10\)), we need to find two numbers that multiply to \(a\times c=3\times(- 10)=- 30\) and add up to \(b=-13\)

The numbers are \(-15\) and \(2\) since \(-15\times2=-30\) and \(-15 + 2=-13\)

We rewrite the middle term using these two numbers:

\(3x^{2}-15x+2x - 10\)

Now we factor by grouping:

Group the first two terms and the last two terms:

\((3x^{2}-15x)+(2x - 10)\)

Factor out the GCF from each group:

\(3x(x - 5)+2(x - 5)\)

Now we can factor out \((x - 5)\):

\((3x + 2)(x - 5)\)

So the factored form of the quadratic is \((3x + 2)(x - 5)\)

Step3: Write the fully factored form of the polynomial

Putting it all together, the factored form of \(f(x)\) is:

\(f(x)=-2x(3x + 2)(x - 5)\)

Step4: Find the roots

To find the roots, we set \(f(x) = 0\):

\(-2x(3x + 2)(x - 5)=0\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\)

• 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:

The roots of the polynomial \( f(x)=-6x^{3}+26x^{2}+20x \) are \( x = 0\), \( x=-\frac{2}{3}\) and \( x = 5\)