# Thread: Expanding triple brackets (easy?)

1. ## Expanding triple brackets (easy?)

I'm gonna feel really dumb asking this but how do you expand triple brackets?

I can do single and double brackets no problem with the FOIL method, but what multiplies with what when you have triple brackets?

Also I was hoping you could help me on a question also here it is:

$
x^3 - 2x^2 - 11x + 12 = 0$

I've deduced that $(fx = 1)$ because $1^3 - 2 (1^2) - 11 (1) + 12 = 0
$

But what do I do next to factorise?

2. Hello,

Originally Posted by Nick87
I'm gonna feel really dumb asking this but how do you expand triple brackets?

I can do single and double brackets no problem with the FOIL method, but what multiplies with what when you have triple brackets?

Also I was hoping you could help me on a question also here it is:

$
x^3 - 2x^2 - 11x + 12 = 0$

I've deduced that $(fx = 1)$ because $1^3 - 2 (1^2) - 11 (1) + 12 = 0
$

But what do I do next to factorise?

If $\alpha$ is a root of a polynomial $P(x)$ of degree n, then you can write : $P(x)=(x-\alpha)Q(x)$, with $Q(x)$ a polynomial of degree n-1.

Here, you can write that $x^3-2x^2-11x+12=(x-1)Q(x)$, where $Q(x)$ is of degree 2 --------> $Q(x)=ax^2+bx+c$

Try to find a,b & c

3. $x^3-2x^2-11x+12)\div(x-1)=x^2-x-12$

Now factor the quadratic $x^2-x-12=0$

$(x-4)(x+3)=0$

$x=4 \ or \ x=-3$

Solutions {-3, 1, 4}

Give an example of your triple bracket problem.

4. Example of my triple bracket problem?

I just need to PROVE that $(x-1)(x-3)(x+2)$ = $x^3 - 2x^2 - 5x + 6$

I know that it does because of the -3 and 2 being factors making the -2, -5 and +6 but i need to show expansion methods.

By the way thanks Moo and Masters! Muchly helpful! :P

5. First, expand the last two binomials using FOIL, since you know that method.

$(x-1)(x^2-x-6)$

Then, use the distributive property to multiply the binomial and the trinomial:

$x(x^2-x-6)-1(x^2-x-6)$

Continuing:

$x^3-x^2-6x-x^2+x+6$

Combine terms:

$x^3-2x^2-5x+6$

6. Thank you once again. My problem is solved!