1. ## Factoring Help

Are these questions right?

1. 32x^5 - 108x^2
4x^3 (8x^2 - 27)

2. 8y^6 - 38y^4 - 10y^2
2y^2 (4y^4 - 19y^2 - 5)
2y^2 (y^2 - 5)(4y^2 + 1)

Also, how would I solve:

z^5 - 32z^4 - 16z + 48

My teacher did not explain to us how to do this...

2. ## Re: Factoring Help

Originally Posted by callmebea
Are these questions right?

1. $\displaystyle 32x^5 - 108x^2$
$\displaystyle 4x^3 (8x^2 - 27)$
No, there's an error here.

2. $\displaystyle 8y^6 - 38y^4 - 10y^2$
$\displaystyle 2y^2 (4y^4 - 19y^2 - 5)$
$\displaystyle 2y^2 (y^2 - 5)(4y^2 + 1)$
Looks good.

Also, how would I solve:

$\displaystyle z^5 - 32z^4 - 16z + 48$

My teacher did not explain to us how to do this...
$\displaystyle z^5 - 32z^4 - 16z + 48$
This doesn't factor. Are you sure this is exactly the polynomial?

3. ## Re: Factoring Help

Could you give me a clue as to what the error I made is?

4. ## Re: Factoring Help

If you multiply out the factored expression you arrived at, you'll see that the result does not match the polynomial you were given in the problem. That will show you where the error is.

5. ## Re: Factoring Help

Originally Posted by callmebea
Are these questions right?

1. 32x^5 - 108x^2
4x^3 (8x^2 - 27)

2. 8y^6 - 38y^4 - 10y^2
2y^2 (4y^4 - 19y^2 - 5)
2y^2 (y^2 - 5)(4y^2 + 1)

Also, how would I solve:

z^5 - 32z^4 - 16z + 48

My teacher did not explain to us how to do this...
1. The HCF is 4x^2, not 4x^3...

2. Correct

3. You are expected to know that if you can find a z = a such that f(a) = 0, then (z - a) is a factor. You are also expected to know that a can only be factors of 48...
Once you have found a factor (z - a), long divide, follow the same process again with the remaining factor until it's completely factorised.