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

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!

My answer:

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

/

[(x+3)x^2]^2

My tutor's simplified answer:

-6x^3 -12x^2 - 13x -42

/

(x+3)^2 x^3

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:

Distribute and get:

Distribute some more and get:

Combine like terms and get:

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

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!

My answer:

(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

My tutor's simplified answer:

-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:

Your tutor's answer is:

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.

**Youkla**'s answer is correct.

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

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

WHY are you reactivating old threads, Mr Spam?

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

my apologies. I didn't realize this pushes the threads up. I'm new to this place. I will stop replying to old threads.