Factor: c^3+64

See factoring pattern on attachment.

c^3+64 is the sum of two cubes. c³ + 4³

c³ + 4³ = (c + 4) ( ? - ? + ? )

(c^2-4c+16)

Yes. c³ + 4³ = (c + 4)* (c^2-4c+16)

Can you help my factor this? \[z^4-21z^2\]

Factor out the largest common factor of z². z² ( ? - ?)

(z^2-7)?

z^4 - 21 z^2 = z² (z² - 21) is what I got.

Yours multiplies back to z^4 - 7 z^2 which is not the original problem.

Oh i see now. thank you! :D

I think it is a good idea to check the factoring, especially the monomial ones, by multiplying back.

Do you have another problem?

yes.

Factor out the GCF x+6+3a(x+6)

This is factoring by grouping.

x+6+3a(x+6) = 1* (x + 6) + 3a * ( x + 6 )

( x + 6 ) is the common factor.

so it would be (3a+1)(x+6)?

It is like having 6x + 6y and factoring out the common factor of 6. But, the common factor is a binomial this time.

1* (x + 6) + 3a * ( x + 6 ) = ( x + 6 ) * ( 1 + 3a) which is the same answer you got.

Next question :)

t^2+10ty+16y^2

I don't see any common monomial factors.

so that would make it Prime right?

It is not prime.

1*t^2 + 10ty + 16 y^2 We will work this by grouping. What are two numbers that multiply to 1*16 and add to 10?

8,2 ?

Correct.

(t+8y)(t+2y)?

1*t^2 + 10ty + 16 y^2 = t^2 + 8 ty + 2 ty + 16y^2.

Yes, that is correct.

Great! :D thank you so much! :D

I am going to make up a problem for you to tackle. Hold on.

Factor 12x^2+16x-35

find the gcf first right?

Look for one, yes.

i can't find one.

Look for two numbers that multiply to (12)*(-35) AND add to 16. This expression will factor.

(6x-7)(2x+5)

I was looking a prime factors of 12 and 35. How did you get this: (6x-7)(2x+5)

6*2 =12 giving 12x^2 6*5=30 -7*2=-14 -7*5=-35 12x^2+30x-14x-35 =12x^2+16-35

Correct. You are showing that the factorization is correct. The more factoring problems on which you practice, the better you will get. Consider cranking out the problems on the site http://library.thinkquest.org/29292/quadratic/2factoring/practice.htm The answers are on the second page.

How about this problem: Factor: (x^2 - 1) - ( -1 - x)

This is factoring by grouping.

(x^2 - 1) - 1* ( -1 - x) = (x^2 - 1) + 1 + x. Use the distributive property to multiply the "understood" -1 by (-1 - x)

= (x^2 - 1) + (1 + x). What do you know about the factorization of this: (x^2 - 1)

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