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.
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:
$\displaystyle 72x^2-2y^2=2(36x^2-y^2)={\color{red}2(6x-y)(6x+y)}$
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)
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!!
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?
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.
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 I am so slow at factorising stuff like that.