# How do I simplify from my answer to the tutor's answer?

• Oct 21st 2011, 08:48 AM
econolondon
Hi

There's an algebra problem I've been working on - I've got as far as I can. My tutor says my answer is correct but can be simplified down to his answer - but I can't see how. If anyone can explain the steps on how to simplify to the answer I'd be really grateful!

(x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]
/
[(x+3)x^2]^2

-6x^3 -12x^2 - 13x -42
/
(x+3)^2 x^3
• Oct 21st 2011, 11:01 AM
Youkla
Re: How do I simplify from my answer to the tutor's answer?
Unless you wrote it down incorrectly or I translated your heiroglyphics incorrectly, I got a different answer:

$\frac{2x^2(x+3) - (2x+7)[2x(x+3) + x^2]}{[x^2(x+3)]^2}$

Distribute and get:

$\frac{2x^3 + 6x^2 - (2x+7)(3x^2 + 6x)}{[x^2(x+3)]^2}$

Distribute some more and get:

$\frac{2x^3 + 6x^2 - 6x^3 - 12x^2 - 21x^2 - 42x}{[x^2(x+3)]^2}$

Combine like terms and get:

$\frac{-4x^3 - 27x^2 - 42x}{x^4(x+3)^2}$

Factor an x out of the numerator and cancel with an x in the denominator and get:

$\frac{-4x^2 - 27x - 42}{x^3(x+3)^2}$
• Oct 21st 2011, 11:04 AM
SammyS
Re: How do I simplify from my answer to the tutor's answer?
Quote:

Originally Posted by econolondon
Hi

There's an algebra problem I've been working on - I've got as far as I can. My tutor says my answer is correct but can be simplified down to his answer - but I can't see how. If anyone can explain the steps on how to simplify to the answer I'd be really grateful!

(x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]
/
[(x+3)x^2]^2

{(x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]} / [(x+3)x^2]^2

-6x^3 -12x^2 - 13x -42
/
(x+3)^2 x^3

(-6x^3 -12x^2 - 13x -42)/[(x+3)^2 x^3]

In LaTeX , your answer is: $\frac{(x+3)x^2(2) - (2x+7)[(x+3)2x + x^2]}{((x+3)x^2)^2}$

Your tutor's answer is: $\frac{-6x^3 -12x^2 - 13x -42}{(x+3)^2 x^3}$

Factor x out of the numerator. Cancel with a factor of x in the denominator.

Expand (multiply out) all of the terms in the numerator, then combine like terms .

There should not be an x^3 term in the numerator after cancelling. The tutor's answer is in error.