How would I factor these polynomials completely?

**deathtolife04** · August 29th 2007, 07:08 PM

I was supposed to factor $16p^2-64$ , and following the steps I ended up with $8(2p^2-8)$ I don't think that's factored COMPLETELY though?

Then I was supposed to factor $4n^2-32n+48$ , and I got $4(n^2-8n+12)$ . I don't think that's factored completely either?! Were there different steps I was supposed to follow or something?

your input would be greatly appreciated!
thanks!

**Jhevon** · August 29th 2007, 07:25 PM

Originally Posted by deathtolife04

$16p^2-64$

Hint: this is the difference of two squares, there's a formula for factoring such expressions, do you know it?

$4n^2-32n+48$ , and I got $4(n^2-8n+12)$

well, you are right for feeling that way, it isn't complete. have you ever heard of "foiling"?

**deathtolife04** · August 29th 2007, 07:32 PM

Originally Posted by Jhevon

Hint: this is the difference of two squares, there's a formula for factoring such expressions, do you know it?

well, you are right for feeling that way, it isn't complete. have you ever heard of "foiling"?

I don't think I'm familiar with a formula for factoring expressions like the first one.

Yes, I've heard of foiling. I think it may be starting to come back to me, how to turn them into 'foil' problems. Is that the thing where I would need to find two numbers whose product is 48, and the sum would have to be -32? And then you somehow turn that into a 'foil' problem? I can't remember exactly...

**Jhevon** · August 29th 2007, 07:39 PM

Originally Posted by deathtolife04

I don't think I'm familiar with a formula for factoring expressions like the first one.

here's the formula. try it and tell me what you get.

$x^2 - y^2 = (x + y)(x - y)$

Yes, I've heard of foiling. I think it may be starting to come back to me, how to turn them into 'foil' problems. Is that the thing where I would need to find two numbers whose product is 48, and the sum would have to be -32? And then you somehow turn that into a 'foil' problem? I can't remember exactly...[/quote]you already made it simpler didn't you? by factoring out the 4? don't worry about 48 and 32, now all you have to worry about is 8 and 12. (By the way, if you are factoring the original expression, it would be even harder, since the coeffcient of $n^2$ is not 1. with experience you should be able to foil it, but beginners have this whole process they have to go through to factor by groups and stuff.

anyway, you were right so far, now you just have to put it together. let's say the two numbers you found were $a$ and $b$ (and they can be positive or negative), then the answer would be in the form:

$(n + a)(n + b)$

contracting the original expression into that form is called "foiling." try it

This is my 31

th post!!!!

**deathtolife04** · August 29th 2007, 07:48 PM

ok, so for the first one would that be: $16(p-2)(p+2)$ ?

I'm still working on the second one...

**Jhevon** · August 29th 2007, 08:00 PM

Originally Posted by deathtolife04

ok, so for the first one would that be: $16(p-2)(p+2)$ ?

that's fine. i would probably say $(4p + 8)(4p - 8)$ though. it looks neater that way, there's no constant hanging out in front, but to each his own

**deathtolife04** · August 29th 2007, 08:26 PM

Originally Posted by Jhevon

that's fine. i would probably say $(4p + 8)(4p - 8)$ though. it looks neater that way, there's no constant hanging out in front, but to each his own

ok, yeah, that sounds better

so for the second one I figured out that -6 and -2 have a product of 12 and a sum of -8

so, how would i write my final answer? i think if i just write it 4(n-6)(n-2) that would be wrong? cuz then it would look like you just had to distribute the 4 to the n-6?????

**Jhevon** · August 29th 2007, 08:29 PM

Originally Posted by deathtolife04

ok, yeah, that sounds better

so for the second one I figured out that -6 and -2 have a product of 12 and a sum of -8

so, how would i write my final answer? i think if i just write it 4(n-6)(n-2) that would be wrong? cuz then it would look like you just had to distribute the 4 to the n-6?????

write (2n - 12)(2n - 4)

i split the 4 into 2 times 2, distributed one 2 in the first set of brackets and the other in the second. it is what you would get if you foiled directly

on second thought, i think it may be better if you keep the constants out in front, it looks messier, but it is factored "completely" when in that form

**deathtolife04** · August 29th 2007, 08:30 PM

ok, thanks!

**deathtolife04** · August 29th 2007, 08:50 PM

Originally Posted by Jhevon

write (2n - 12)(2n - 4)

i split the 4 into 2 times 2, distributed one 2 in the first set of brackets and the other in the second. it is what you would get if you foiled directly

on second thought, i think it may be better if you keep the constants out in front, it looks messier, but it is factored "completely" when in that form

actually, so would i need to put a 4 in front of each set of parenthesis? or keep it as 4(n-6)(n-2)
?

**Jhevon** · August 29th 2007, 08:51 PM

Originally Posted by deathtolife04

$6p^2 + p - 12$

foil

Originally Posted by deathtolife04

actually, so would i need to put a 4 in front of each set of parenthesis? or keep it as 4(n-6)(n-2)
?

yes, do what you did before

you got the other one ok?

**deathtolife04** · August 29th 2007, 09:06 PM

Originally Posted by Jhevon

yes, do what you did before

you got the other one ok?

yeah, i think i'm good now! thanks

let me clarify, though. do what i did before as in, keep it with ONE 4, right?

so: 4(n-6)(n-2)
but then wouldn't they think that the 4 should only be distributed through the FIRST set of parenthesis? or does it matter?

**Jhevon** · August 29th 2007, 09:11 PM

Originally Posted by deathtolife04

yeah, i think i'm good now! thanks

let me clarify, though. do what i did before as in, keep it with ONE 4, right?

so: 4(n-6)(n-2)
but then wouldn't they think that the 4 should only be distributed through the FIRST set of parenthesis? or does it matter?

to "factor completely" means we should be able to "pull anything else out." if you distribute the 4, then it will become something you can "pull out"

so leave it as it was the first time, which is what i believe you have here

**deathtolife04** · August 29th 2007, 09:14 PM

Originally Posted by Jhevon

to "factor completely" means we should be able to "pull anything else out." if you distribute the 4, then it will become something you can "pull out"

so leave it as it was the first time, which is what i believe you have here

ok. thanks.

Thread: How would I factor these polynomials completely?

LinkBack

Thread Tools

Display

How would I factor these polynomials completely?

Similar Math Help Forum Discussions

Factor Completely

Factor out COMPLETELY

Factor completely

Factor completely

factor this completely

Search Tags