# Thread: Factorising with Factor Theorem

1. ## Factorising with Factor Theorem

Hi there, any help on these two questions would be muchly appreciated. Im trying to use Factor Theorem to answer a few questions, I've been successful up until these last two questions.

Factorise:

1) $2x^3+7x^2-7x-12$

and

2) $x^3-3x^2-4x+12$

Thanks.

2. For the first one, you could try dividing it by some possible roots, like -1, -4, or 3/2. Try dividing by (x+1), (x+4) or (2x-3). If it reduces to a quadratic by dividing by either of those, then you are in business.

You could also rewrite this one as
...............7x^2..........-7x
$2x^{3}+\boxed{10x^{2}-3x^{2}}+\boxed{8x-15x}-12$

Factor: $2x(x^{2}+5x+4)-3(x^{2}+5x+4)$

$(2x-3)(x^{2}+5x+4)=(2x-3)(x+4)(x+1)$

For the second one, same logic as before. Try dividing this one by (x-3), x-2, or x+2. It should reduce to a quadratic and be easy thereafter. Okey-doke?.

3. Thanks for that! it really helps!

4. Hello, Nick87!

The Factor Theorem has two powerful features:

Given a polynomial $P(x)$

if $P(a) = 0$, then $(x-a)$ is a factor of $P(x).$

The only rational zeros of $P(x)$ are of the form $\frac{n}{d}$
. . where $n$ is a factor of the constant term and $d$ is a factor of the leading coefficient.

I'm using the Factor Theorem.
I've been successful up until these last two questions.

$1)\;\;P(x) \;=\;2x^3+7x^2-7x-12$

The factors of the constant term are: . $\pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12$
The factors of the leading coefficient are: . $\pm1,\:\pm2$
The possible rational zeros are: . $\pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12,\:\pm\fra c{1}{2},\:\pm\frac{3}{2}$

We find that: . $P(1) \;=\;2\!\cdot\!1^3 + 7\!\cdot\!1^2 - 7\!\cdot\!1 - 12 \;=\;-10\;\neq\;0$

But: . $P(\text{-}1) \;=\;2(\text{-}1)^3 + 7(\text{-}1)^2 - 7(\text{-}1) - 12 \;=\;0$

. . Since $x = \text{-}1$ is a zero, then $(x+1)$ is a factor.

Using long (or synthetic division): . $P(x) \;=\;(x+1)(2x^2 + 5x- 12)$

. . Therefore: . $P(x) \;=\;(x+1)(x+4)(2x-3)$

$2)\;\;P(x)\;=\;x^3-3x^2-4x+12$

The possible zeros are: . $\pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12$

We find that: . $P(2) \;=\;2^3 - 3\!\cdot\!2^2 - 4\!\cdot\!2 + 12 \;=\;0$

. . Hence, $(x-2)$ is a factor.

. . and we have: . $P(x) \;=\;(x-2)(x^2-x-6)$

Therefore: . $P(x) \;=\;(x-2)(x+2)(x-3)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This one could have been factored "by grouping" . . .

. . $x^3 - 3x^2 - 4x + 12 \;=\;x^2(x-3) - 4(x-3)
$

. . $= \;(x-3)(x^2-4) \;=\;(x-3)(x-2)(x+2)$

5. Wow, thanks for this help guys! I now understand Factor Theorem a lot more

Galactus, Soroban I salute!