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The entire problem is 2x^3+16/6x+12x times 5/x^2-2x+4

2006-11-08 07:17:28 · 5 answers · asked by EmotionalStrength1022 2 in Science & Mathematics Mathematics

5 answers

2x³ + 16 = 2(x³ + 8)
= 2(x³ + 2³)
= 2(x + 2)(x² - 2x + 4)

So (2x³+16)/(6x + 12) * 5/(x² - 2x + 4)
= 2(x + 2)(x² - 2x + 4)/6(x + 2) * 5/(x² - 2x + 4)
=5/3 (on simplifying)

2006-11-08 07:50:08 · answer #1 · answered by Wal C 6 · 1 1

Divide/factor out the 2:

2( x^3 + 8)

That's the only thing you can do to it all by itself. Nothing else is common to factor.

However, looking at the entire problem, there is a lot more to do. First, look at each piece:

(2x^3 + 16) ------> the 2 is common in all terms, so it can be
--------------- ------> divided out of everything
(6x + 12)

(x^3 + 8)
------------
(3x + 6)

Factor the denominator further:
(x^3 + 8)
----------- or 1/3*((x^3 + 8)/(x + 2))
3(x + 2)

There is nothing else in the problem that can simplified, so now you will have to do the multiplication. You should end up with this product:

5(x^3 + 8)
--------------
3(x^3 + 8) -----> all the other terms cancel out

Ending up with 5/3!

Now I'm going to spend some time trying to figure out if there is a way to know that x^3 + 8 = (x + 2)(x^2 -2x + 4). It will probably drive me nuts for the rest of the afternoon until I can get home and look at my algebra book! ;-)

2006-11-08 15:41:26 · answer #2 · answered by swimmerd76 2 · 1 1

Factoring sum of cubes;

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

Your first problem;

2x^3+16

Factor out 2;

2(x^3+8)

cube root of 8 is 2; let x=x and y=2

2(x+2)(x^2-2x+4)

The rest of the problem is

(numerator * numerator) / (denominator * denominator)

(5(2)(x+2)(x^2-2x+4)) / (18x(x^2-2x+4))

Cancel out 2 and (x^2-2x+4);

(5(x+2))/(9x)

First problem is 2(x+2)(x^2-2x+4)

Entire problem is (5x+10)/(9x)

--------- --------- --------- --------- --------- --------- ---------
UPDATE !!!
--------- --------- --------- --------- --------- --------- ---------

p.s. if your first rational denominator is supposed
to be 6x^2+12x then the entire problem is;

Factor out 6x;

6x(x+2)

Then (numerator * numerator) / (denominator * denominator);

(5(2)(x+2)(x^2-2x+4)) / (6x(x+2)(x^2-2x+4))

Cancel out 2, (x+2), and (x^2-2x+4)

(5)/(3x)

Entire problem is (5)/(3x)

2006-11-08 16:02:56 · answer #3 · answered by Chris 5 · 0 0

you solve for x. do everything else first and then isolate the x^ and then you should be able to figure it out by knowing what the number equals.

2006-11-08 15:20:37 · answer #4 · answered by Sharp Marble 6 · 0 1

hmm
2x+2 (x-4) (x-2)
no that ain't quite right

2006-11-08 15:22:35 · answer #5 · answered by mike c 5 · 0 1

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