# just checking that I got these right

Printable View

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Dec 12th 2008, 12:37 PM
tpoma
just 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 PM
chella182
Quote:

Originally Posted by tpoma
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!

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 PM
tpoma
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 PM
masters
Quote:

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

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

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:

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

Quote:

Originally Posted by tpoma
$\displaystyle 72x^2 - 2y^2$ should = $\displaystyle (2y-3x) (2y-36x)$ right?

Thanks so much!

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 PM
tpoma
Quote:

Originally Posted by masters
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:

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

I wish I had your brains!!! This is so fustrating to me!! Thank you once again! I will get it one day hopefully lol.....(Crying)
• Dec 12th 2008, 01:08 PM
masters
Quote:

Originally Posted by tpoma
I wish I had your brains!!!

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 PM
tpoma
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 PM
chella182
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 PM
tpoma
Quote:

Originally Posted by chella182
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.

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 PM
masters
Quote:

Originally Posted by tpoma
Ok got one more for you, and hopefully the last one....lol

$\displaystyle c^2 - 3cd - 18d^2$

is this correct?

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

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 PM
chella182
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 PM
tpoma
OH brother! You make that look so easy! (Giggle) Thanks!
• Dec 12th 2008, 02:34 PM
tpoma
Last 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 PM
masters
Quote:

Originally Posted by tpoma
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)

Yes!
• Dec 13th 2008, 09:35 PM
great_math
Quote:

Originally Posted by tpoma
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) god I hope so!!

you can check my multiplying it again if you are not sure..... that's the way to check yourself
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last