Taylor G.

asked • 02/05/16

More factoring!!!

How do I get from 4x2-32x to 4x(x-2)(x2+2x+4)???

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LaToshia L. answered • 02/05/16

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That problem is not correct because 4x(x-2) gives you 8x2-8
 
8x2-8(x2+2x+4) gives you 8x4, which was not a solution you provided.  Can you double check the problem again?
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Taylor G.

I've multiple problems in the notes section of my class, this is one of the practice probelms. Thank you!
 
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02/05/16

Elwyn D. answered • 02/05/16

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You don't.
4x^2 - 32x = 4x (x-8)
 
 
If you miscopied, and the problem is actually 4x^4 - 32x, then
4x^4 - 32x = 4x (x^3 -8)
 
x^3 -8 is the difference of two cubes, which is a formula you should simply memorize
a^3 - b^3 = (a - b) (a^2 + ab +b^2)
 
so,
 
4x (x^3 - 8) = 4x(x-2)(x^2 + 2x + 4)
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Elwyn D.

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You should also memorize the sum of two cubes:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
 
Problems containing the sum or difference of two cubes are very common.  Factoring them anew each time would be time-consuming, frustrating, and possibly a source of error.  Memorize those formulas: save time and get it right.
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02/05/16

Taylor G.

How do I work that formula?
 
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02/05/16

Elwyn D.

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Look for the pattern of cube + cube or cube - cube.

Let's select an ornate example:
27x^6y^3 - 8a^9b^12

since every component of 27x^6y^3 is a cube 27 = 3^3, x^6 = (x^2)^3, and y^3 (obviously), the term 27x^6y^3 = (3 x^2 y)^3
(3 x^2 y)= a

8 a^9 b^ 12 = (2 a^3 b^4)^3 
(2 a^3 b^4) = b

so, 27x^6y^3 - 8a^9b^12 = (3 x^2 y)^3 - (2 a^3 b^4)^3 
 
(a-b)(a^2 + ab + b^2) =
(3 x^2 y - 2 a^3 b^4)[(3 x^4 y^2) + 6 x^2 y a^3 b^4)(4 a^6 b^8)]
 
You can solve a lot of problems by looking for the patterns of difference of two cubes or sum of two cubes.  They crop up a lot.  I promise that they even occur outside of math-textbook exercises designed to drill the skill home.
 
Memorize the formulas and look for the patterns.
 
 
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02/05/16

Elwyn D.

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I may be repeating myself. Wyzant seems to have misplaced my comment. If it is just slow in posting I apologize for repeating myself.
 
Look for the patterns Difference of Two Squares, Sum of Two Squares.
 
27x^3y^6 - 8a^9b^12 is the difference of two squares
(3xy^2)^3 - (2a^3b^4) so, it factors to
(3xy^2 - 2a^3b^4)(9x^2y^4 + 6xy^2a^3b^4 + 4a^6b^8)
 
It works, and it is not hard to recognize the pattern within a problem.
BTW difference of cubes and sum of two cubes problems are not just the province of math-texts, you will actually encounter them in the physical sciences.
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02/05/16

Elwyn D.

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This is my third attempt to post an answer for you.  I apologize but my explanations are getting shorter.
 
Look for the patterns  cube - cube and cube + cube.  You will find several problems with this pattern in your math book, but they also occur in real life in the physical sciences often enough to warrant  your attention.
 
The expression 27a^3 - 8b^6 is the difference of two cubes 27a^3 = (3a)^3 and 8b^6 = (2b^2)^3
so, a = 3a and b = 2b^2
 
so, it can be factored (3a-2b^2)(9a^2 + 6ab^2 + 4b^4)
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02/05/16

Elwyn D.

tutor
Sorry, that did NOT come out right. 
 
I have tried twice to answer your reasonable and welcome question "How do I work that formula".  The two other attempts have disappeared.  They were not saved.
 
I was NOT expressing impatience with your question.
 
Hope my answer made sense.  Please, have no qualms about commenting back if you still need clarification.
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02/05/16

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