Factoring Binomials

For college beginning algebra.

~Please explain every step because I'm mathmatically stupid~


Problem:

5k^3+40


The answer is SUPPOSIDLY 5(K+2)(K^2-2K+4)

but i dont see how they get it.




Here's the example they put in the book for this type of problem.


Factor:
64x^3 + 1

Solution:
(4x)^3 + 1^3
=
(4x+1)[(4x)^2-(4x)(1)+1^2]
=
(4x+1)(16x^2-4x+1)


and so from what im takin it is..
if you have
5k^3+40

Then first step is.. you gotta make the 40 into a cubed like the 5k^3..

but.. as far as i know, NOTHING cubed is 40. I had my dad look at this too and he has no idea what to do.. and neither do i.


Thanks

~Cait

Quote:

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Originally Posted by Cait

For college beginning algebra.

~Please explain every step because I'm mathmatically stupid~


Problem:

5k^3+40


The answer is SUPPOSIDLY 5(K+2)(K^2-2K+4)

but i dont see how they get it.




Here's the example they put in the book for this type of problem.


Factor:
64x^3 + 1

Solution:
(4x)^3 + 1^3
=
(4x+1)[(4x)^2-(4x)(1)+1^2]
=
(4x+1)(16x^2-4x+1)


and so from what im takin it is..
if you have
5k^3+40

Then first step is.. you gotta make the 40 into a cubed like the 5k^3..

but.. as far as i know, NOTHING cubed is 40. I had my dad look at this too and he has no idea what to do.. and neither do i.


Thanks

~Cait

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ok, first you need to know the rule:

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

now to do the problem:

5k^3+40 .........first thing to do is factor out the 5

=> 5(k^3 + 8) ..............now 8 = 2^3
=> 5(k^3 + 2^3) ............now apply the rule above
=> 5(k+2)(k^2 - 2k + 2^2) = 5(k+2)(k^2 - 2k + 4)

awesome.
that worked.. and also for other probs.

which leads me to another question! yay.

How come they want you to split the number 8 into 2 and 4 in the problem

8a^3+1

Ans: (2a+1)(4a^2-sa+1)

and not split the 8 in this problem?

8r^3-64

Ans: 8(r-2)(r^2+2r+4)


It's prolly because it has a +1 at the end insted of.. say.. 3 and 2..?? right??


there's so many particlar rules about the tiniest things and i will never remember them :|

Quote:

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Originally Posted by Cait

awesome.
that worked.. and also for other probs.

which leads me to another question! yay.

How come they want you to split the number 8 into 2 and 4 in the problem

8a^3+1

Ans: (2a+1)(4a^2-sa+1)

and not split the 8 in this problem?

8r^3-64

Ans: 8(r-2)(r^2+2r+4)


It's prolly because it has a +1 at the end insted of.. say.. 3 and 2..?? right??


there's so many particlar rules about the tiniest things and i will never remember them :|

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Quote:

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8a^3+1

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remember 8 = 2^3
also remember that x^a * y^a = (xy)^a
now to do the problem

8a^3 + 1
= 2^3 * a^3 + 1
= (2a)^3 + 1^3
= ((2a) + 1)((2a)^2 - (2a)(1) + 1^2)
= (2a + 1)(4a^2 - 2a + 1)

Quote:

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8r^3-64

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two ways to do this one. one way is the same way we did 5k^3 + 40, another way is the way i did the problem before this one.

First Method:

8r^3 - 64
= 8(r^3 - 8)
= 8(r^3 - 2^3)
= 8(r - 2)(r^2 + 2r + 2^2)
= 8(r - 2)(r^2 + 2r + 4)

Second Method:

8r^3 - 64
= 2^3 * r^3 - 4^3
= (2r)^3 - 4^3
= ((2r) - 4)((2r)^2 + 8r + 4^2)
= (2r - 4)(4r^2 + 8r + 16) ...........factor 2 out of the first group and 4 out of the second, we get

= 2(r - 2) * 4(r^2 + 2r + 4) ........now multiply the 4 and the 2 and put them in front
= 8(r - 2)(r^2 + 2r + 4)

Quote:

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Originally Posted by Cait

there's so many particlar rules about the tiniest things and i will never remember them :|

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if you keep thinking like that, you won't

how come i remember them?

Alright, TY again. :3

Quote:

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how come i remember them?

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Cuz you're smart! :D



I got 2 sections left.. Hopefully I wont have any trouble with them..

Quote:

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Originally Posted by Cait

Alright, TY again. :3



Cuz you're smart! :D



I got 2 sections left.. Hopefully I wont have any trouble with them..

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not only am i not smart, but my memory is horrible, which brings me back to the advice i gave you in another thread

Quote:

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it gets easier with practice and a good attitude, do a lot of problems and smile and be happy about it:D

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