# How do you factor 3x^2( 4x-12)^2 + x^3(2)(4x-12)(4)?

Oct 15, 2015

$8 {x}^{2} \left(x - 3\right) \left(5 x - 7\right)$
This is more of a guide to approach than deriving the correct answer. You can check that!!!!!

#### Explanation:

The trick is to look for common elements that can be factored out and then playing around until you find the solution.
Looking at the question you will observe that ${x}^{2} \left(4 x - 12\right)$ is a common factor. So let us 'play' with that and see what we get.
I am taking it one step at a time so that you can see the process and develop your own faster approaches.

Consider $3 {x}^{2} {\left(4 x - 12\right)}^{2}$
this can be factored so that you have:

${x}^{2} \left(4 x - 12\right) \times 3 \left(4 x - 12\right)$ ..............................( 1 )

Consider ${x}^{3} \left(2\right) \left(4 x - 12\right) \left(4\right)$
This can be factored so that you have:

${x}^{2} \left(4 x - 12\right) \times 8 x$.............................................( 2 )

Putting ( 1 ) and ( 2 ) together gives:

${x}^{2} \left(4 x - 12\right) \times 3 \left(4 x - 12\right) + {x}^{2} \left(4 x - 12\right) \times 8 x$

Factoring again

${x}^{2} \left(4 x - 12\right) \left(3 \left[4 x - 12\right] + 8 x\right)$

${x}^{2} \left(4 x - 12\right) \left(12 x - 36 + 8 x\right)$

${x}^{2} \left(4 x - 12\right) \left(20 x - 36\right)$

Notice that all the numbers in the bracket are even. This means that we can take it 'down' again by factor of 2

$2 {x}^{2} \left(2 x - 6\right) \times 2 \left(10 x - 18\right)$

Again they are all even within the brackets so let us repeat the process:

$4 {x}^{2} \left(x - 3\right) \times 4 \left(5 x - 9\right)$

But $4 {x}^{2}$ from $4 {x}^{2} \left(x - 3\right)$ multiplied by 4 from $4 \left(5 x - 9\right)$ = $8 {x}^{2}$

Substituting this in gives:

$8 {x}^{2} \left(x - 3\right) \left(5 x - 9\right)$

I have not checked that the final answer is correct. I will leave that to you to do!!! However you can see the process types available to you!!!