# Math Help - Factoring Questions - Help!

1. ## Factoring Questions - Help!

We were recently given a worksheet with some problems on it and I'm struggling with a few of the problems. Can anyone lead me in the direction or show me how to do some of these factoring problems?

Factor Fully.

1) 28s^2 - 8st - 20t^2

2) y^2 - (r - n)^2

3) 10x^2 + 38x + 20

4) y^5 - y^4 + y^3 - y^2 + y - 1

I'd greatly appreciate any help given for any of these problems. Thank you!

2. Originally Posted by Jeavus
1) 28s^2 - 8st - 20t^2

2) y^2 - (r - n)^2

3) 10x^2 + 38x + 20

4) y^5 - y^4 + y^3 - y^2 + y - 1
1) $28s^2-8st-20t^2=4(7s^2-2st-5t^2)$, it remains to factorice $7s^2-2st-5t^2=7s^2-(7st-5st)-5t^2$, can you take it from there?

2) Use $a^2-b^2=(a+b)(a-b)$

3) Similar as 1)

4) Factorice by grouping

3. Originally Posted by Jeavus
We were recently given a worksheet with some problems on it and I'm struggling with a few of the problems. Can anyone lead me in the direction or show me how to do some of these factoring problems?

Factor Fully.

1) 28s^2 - 8st - 20t^2
$28s^2 - 8st - 20t^2$

First rule, factor any constant out of the whole thing that you can:
$= 4(7s^2 - 2st - 5t^2)$

Now, since 7 and 5 are prime all we need to do is arrange the 7 and 5 in some way to multiply out correctly. Trial and error gives me:
$= 4(7s + 5t)(s - t)$

Originally Posted by Jeavus
2) y^2 - (r - n)^2
Remember $a^2 - b^2 = (a + b)(a - b)$?
Let a = y, b = r - n.
$y^2 - (r - n)^2 = (y + (r - n))(y - (r - n)) = (y + r - n)(y - r + n)$

Originally Posted by Jeavus
3) 10x^2 + 38x + 20
$10x^2 + 38x + 20$

$= 2(5x^2 + 19x + 10)$
This can't be factored any further than this.

Originally Posted by Jeavus
4) y^5 - y^4 + y^3 - y^2 + y - 1
This is a fun one, isn't it? Factor by grouping:
$y^5 - y^4 + y^3 - y^2 + y - 1$

$= (y^5 - y^4 + y^3) + (- y^2 + y - 1)$

$= y^3(y^2 - y + 1) - (y^2 - y + 1)$

Notice that both terms in parenthesis are the same, so we can factor again:
$= (y^3 - 1)(y^2 - y + 1)$

Now, the second factor can't be factored any further, but the first term can:
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

So
$y^5 - y^4 + y^3 - y^2 + y - 1 = (y^3 - 1)(y^2 - y + 1) = (y - 1)(y^2 + y + 1)(y^2 - y + 1)$

-Dan

4. 1) $28s^{2} - 8st - 20t^{2}$
The trick here is to factor out 4 first.

Then we have $4(7s^{2}-2st-5t^{2})$

Now, look at the middle term in the parentheses.

Rewrite -2st as 5st-7st:

$4(7s^{2}+5st-7st-5t^{2})$

$4((7s^{2}+5st)-(7st+5t^{2}))$

$4(s(7s+5t)-t(7s+5t))$

$4(s-t)(7s+5t)$

5. Thanks so much for all the help! That all made perfect sense.

I have a few more questions that I am struggling with if you'd be so kind as to direct me in the correct path:

1) 3v^2 - 11v - 10

Personally I don't think this one can be factored, as nothing multiplies to (-30) and adds to (-11).

2) h^3 + h^2 + h + 1

3) 9(x + 2y + z)^2 - 16(x - 2y + z)^2

All help will be greatly appreciated.

6. They all have the same idea, try it by your own!

7. I'm completely stumped with the first one. =/

I've tried various grouping methods with the second.

And I'm assuming there is some kind of trinomial factoring but again, I'm lost as to finding the light.

Any hints? :|

8. Here's where I'm stuck for the 2nd one:

h^3 + h^2 + h + 1
= h(h^2 + h + 1) + 1

Now I need two numbers that multiply to 1, but also add to 1 - which doesn't make sense to me. =/

EDIT AGAIN:

Solved this one!

Anyone have any ideas about the other two?

9. Originally Posted by Jeavus
Thanks so much for all the help! That all made perfect sense.

I have a few more questions that I am struggling with if you'd be so kind as to direct me in the correct path:

1) 3v^2 - 11v - 10
Are you sure that you have the signs right? Could it be:

$v^2-11v+10$?

RonL

10. Yes, I have the signs right.

Un-factorable, right?

11. It has real nasty roots, but you could factorice it finding its roots.

12. Krizalid/anyone else, do you have any hints towards the third one?

9(x + 2y + z)^2 - 16(x - 2y + z)^2

13. Originally Posted by Jeavus
9(x + 2y + z)^2 - 16(x - 2y + z)^2
$9\left( {x + 2y + z} \right)^2 - 16\left( {x - 2y + z} \right)^2 = \left[ {3\left( {x + 2y + z} \right)} \right]^2 - \left[ {4\left( {x - 2y + z} \right)} \right]^2$

Does that make sense?

Now apply the identity $a^2-b^2=(a+b)(a-b)$

14. 9(x + 2y + z)^2 - 16(x - 2y + z)^2
= [3(x + 2y + z)]^2 - [4(x - 2y + z)]^2
a^2 - b^2
a = [3(x + 2y + z)]
b = [4(x - 2y + z)]
(a + b)(a - b)
= ([3(x + 2y + z)] + [4(x - 2y + z)])([3(x + 2y + z)] - [4(x - 2y + z)])

Where do I go from this stage?

15. Originally Posted by Jeavus
Yes, I have the signs right.

Un-factorable, right?
No real factors

RonL

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