Just want to double check that I am doing these right. Not looking for an answer.

5p^4q - 20q^3 should = 5pq(p^3-4q^2) right?

72x^2 - 2y^2 should = (2y-3x) (2y-36x) right?

Thanks so much!

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- Dec 12th 2008, 12:37 PMtpomajust checking that I got these right
Just want to double check that I am doing these right. Not looking for an answer.

5p^4q - 20q^3 should = 5pq(p^3-4q^2) right?

72x^2 - 2y^2 should = (2y-3x) (2y-36x) right?

Thanks so much! - Dec 12th 2008, 12:41 PMchella182
Well the first one is wrong because if you multiply out your answer you get $\displaystyle 5p^{4}q-20pq^{3}$, which isn't what you have on the left-hand side. In other words, you can't take a $\displaystyle p$ outside the bracket.

I don't think the second one is right either because multiplying that out you'd get $\displaystyle 4y^{2}-6xy-72xy+108x^{2}$ or $\displaystyle 4y^{2}-78xy+108x^{2}$ which again isn't what you have on the left-hand side. - Dec 12th 2008, 12:48 PMtpoma
ok re-do

5p^4q - 20q^3 should have been 5q(p^4-4q^2)

Will have to work on the second one..... - Dec 12th 2008, 12:52 PMmasters
Find your greatest common factor over your two terms

$\displaystyle 5p^4q-20q^3$

The common factor is 5q. Factor that out and you have:

$\displaystyle 5q(p^4-4q^2)$

Find your GCF again. That would be 2. Factor that out.

$\displaystyle 72x^2-2y^2=2(36x^2-y^2)$

Inspecting the binomial, you should notice that it is the difference of two squares and factors in this manner:

$\displaystyle a^2-b^2=(a-b)(a+b)$

So, finally we have:

$\displaystyle 72x^2-2y^2=2(36x^2-y^2)={\color{red}2(6x-y)(6x+y)}$

- Dec 12th 2008, 12:56 PMtpoma
- Dec 12th 2008, 01:08 PMmasters

The Brain -- is wider than the Sky --

For -- put them side by side --

The one the other will contain

With ease -- and You -- beside --

The Brain is deeper than the sea --

For -- hold them -- Blue to Blue --

The one the other will absorb --

As Sponges -- Buckets -- do --

The Brain is just the weight of God --

For -- Heft them -- Pound for Pound --

And they will differ -- if they do --

As Syllable from Sound --

*Emily Dickenson (w.1955)* - Dec 12th 2008, 01:38 PMtpoma
Ok got one more for you, and hopefully the last one....lol

c^2 - 3cd - 18d^2

is this correct?

3(cd + 6d^2 - c^2) = 3c(d+2d^2 - c) gee I hope so!! - Dec 12th 2008, 01:48 PMchella182
You can't take a three out because there isn't a 3 in every term (the $\displaystyle c^2$ term). You can't take a $\displaystyle c$ out either because there's no $\displaystyle c$ in the $\displaystyle d^{2}$ term.

- Dec 12th 2008, 01:57 PMtpoma

Ok so.....once again I am not able to do anything with this problem because it is not factorable....there is not a c, d, or a number in each term, so again my answer would be__not able to factor__. Why would my professor put these on here to be factored completely if you can not factor at all? (Headbang) - Dec 12th 2008, 02:04 PMmasters
I hate to burst your bubble, but $\displaystyle 3 \ \ nor \ \ c$ is a common factor of your trinomial expression.

You must factor this as a trinomial. There are different techniques people use to do this, but here I might suggest the "guess and check" method.

If factorable, a trinomial will be the product of two binomials. So, let's start there.

$\displaystyle c^2 - 3cd - 18d^2=(\ \ \ \ \ \ \ \ )(\ \ \ \ \ \ \ \ )$

Now, factoring the first term is easy:

$\displaystyle c^2 - 3cd - 18d^2=(c \ \ \ \ \ \ )(c \ \ \ \ \ \ )$

Now, look at the last term coefficient$\displaystyle -18$. We need to find two factors that multiply together to make $\displaystyle -18$ and add together to make your middle term coefficient $\displaystyle -3$. This is the guess and check part.

Try various factors of $\displaystyle -18$

$\displaystyle -3 \cdot +6 = -18$

$\displaystyle -3 + 6 = 3$

This pair is no good because they don't add to $\displaystyle -3$

$\displaystyle -6\cdot +3 = -18$

$\displaystyle -6 + 3 = -3$

This pair works because they add to $\displaystyle -3$

So, let us finish filling in the two binomial factors. By the way, don't forget that there's a $\displaystyle d^2$ attached to that $\displaystyle 18$ we just factored.

$\displaystyle c^2 - 3cd - 18d^2=(c -6d)(c +3d)$

There you have it. - Dec 12th 2008, 02:05 PMchella182
Actually...

$\displaystyle (c-6d)(c+3d)$

is the answer. If you multiply out those brackets you should get what you started with.

Darn, got beaten to it (Giggle) I am so slow at factorising stuff like that. - Dec 12th 2008, 02:18 PMtpoma
OH brother! You make that look so easy! (Giggle) Thanks!

- Dec 12th 2008, 02:34 PMtpomaLast one I PROMISE!
Last one I PROMISE!

is this correct.....just a plain yes or no answer.

49x^2 + 140x + 100

answer should be......(7x + 10) (7x + 10) right???? Please say yes..lol(Lipssealed) - Dec 12th 2008, 07:38 PMmasters
- Dec 13th 2008, 09:35 PMgreat_math