# Math Help - Factoring

1. ## Factoring

I believe i know the answer this this one, but i just want to make sure.
Could someone tell me the answer to the following:

Factor the difference of two sqaures,
25 - (z+5)²
Also, this one, i am having a hard time finding the answer to:
Completely factor the expression,
5(3-4x)² - 8(3-4x)(5x-1)

Thanks

2. Originally Posted by WilliamR
I believe i know the answer this this one, but i just want to make sure.
Could someone tell me the answer to the following:

Factor the difference of two sqaures,
25 - (z+5)²
Also, this one, i am having a hard time finding the answer to:
Completely factor the expression,
5(3-4x)² - 8(3-4x)(5x-1)

Thanks
In the first, let z+5=a. This makes the problem look a little less distracting. We can always replace the z+5 after the end. So... How would you factor...

$25-a^2$

It would look like this

$(5+a)(5-a)$

now if I put z+5 back in...

$[5+(z+5)][5-(z+5)]$

And you can simplify further.

In the second problem, try putting a=3x-4, and b= 5x-1. Then factor and replace like the last problem.

3. I am still not sure about the answer the the second problem

4. 5(3-4x)² - 8(3-4x)(5x-1)
(3 - 4x) [ 5(3-4x) - 8(5x-1)] ... can you finish?

5. Originally Posted by WilliamR
I am still not sure about the answer the the second problem
Can you factor

$5a^2-8ab$ ?

6. Originally Posted by VonNemo19
Can you factor

$5a^2-8ab$ ?
a(5a-8b)

5(3-4x)² - 8(3-4x)(5x-1)
(3 - 4x) [ 5(3-4x) - 8(5x-1)] ... can you finish?
Erm....240x² -272x+69....?

(3-4x) [ 5(3-4x) - 8(5x-1)]
(3-4x) [ (15 - 20x) - 40x + 8)]
(3-4x) [ -60x + 23 ]

and we are done factoring.

(3-4x) [ 5(3-4x) - 8(5x-1)]
(3-4x) [ (15 - 20x) - 40x + 8)]
(3-4x) [ -60x + 23 ]

and we are done factoring.
I see now, that is clear, thank you .

9. Now substitute back for a and b.

The point that I'm trying to make here is that factors can be a distraction if they contain more than one term like (z+5). But if you look at it like the way that I've showed you, you will become really good really fast.

Sometimes its better not to just give someone a solution....

But actually teach them something.

10. VonNemo19, i agree with you. That is what MATH HELPER FORUM intends to, guiding them. Voila!