Question

What is the best method for fully factoring this polynomial 3x^3-7x^2-6x

Answer 1

To fully factor the polynomial \(3x^3 - 7x^2 - 6x\), we can start by factoring out the greatest common factor (GCF) from each term.

The GCF of the terms \(3x^3\), \(-7x^2\), and \(-6x\) is \(x\). So we can factor \(x\) out:

\[
3x^3 - 7x^2 - 6x = x(3x^2 - 7x - 6)
\]

Next, we need to factor \(3x^2 - 7x - 6\). We can apply the factoring technique for quadratic equations.

To factor the quadratic \(3x^2 - 7x - 6\), we look for two numbers that multiply to \(3 \times -6 = -18\) (the product of the coefficient of \(x^2\) and the constant term) and add up to \(-7\) (the coefficient of \(x\)).

The two numbers that satisfy these conditions are \(-9\) and \(2\) (because \(-9 + 2 = -7\) and \(-9 \cdot 2 = -18\)).

Now we can rewrite \(-7x\) as \(-9x + 2x\):

\[
3x^2 - 9x + 2x - 6
\]

Next, we can group the terms:

\[
(3x^2 - 9x) + (2x - 6)
\]

Factoring each group gives:

\[
3x(x - 3) + 2(x - 3)
\]

Now, we can factor out the common factor \((x - 3)\):

\[
(3x + 2)(x - 3)
\]

Putting it all together, the completely factored form of the original polynomial is:

\[
x(3x + 2)(x - 3)
\]

Thus, the fully factored polynomial is:

\[
\boxed{x(3x + 2)(x - 3)}
\]