# Math Help - Factoring Question

1. ## Factoring Question

I am trying to figure out how to factor out any common factors out of this polynomial: 4(x2+4)3/2(3x+5)1/3 + (3x+5)4/3(x2+4)1/23x
What I thought to do was to look for each base raised to the smallest exponent present in each term. Looking at the polynomial it looks like (3x+5)1/3, and (x2+4)1/2 are the bases raised to the lowest exponent because 3/2 > 1/2, and 4/3>1/3 therefore I tried to factor these out. When I try to do this I end up with 4(x2+4)2/2+(3x+5)3/33x[(3x+5)(x2+4)], and when I multiply the two binomials in the brackets I get 3x3+5x2+12x+20.
The answer in the text is (3x+5)1/3(x2+4)1/2(13x2+15x+16). I don't see how they arrived at this, and I also don't see how you could arrive at the original polynomial through multiplication of the terms in the answer. If someone could tell me what I am doing wrong, and explain how to arrive at the correct answer I would really appreciate it. Thanks a lot, I love this site.

2. ## Re: Factoring Question

There is a mistake in your simplification

It is (x^2+4)^(1/2) (3x+5)^(1/3) { 4(x^2+4) + 3x(3x+5)} This will give you the text answer.

3. ## Re: Factoring Question

Thanks for the reply, however I still don't understand how the bases raised to the 3/2, and 4/3 are less than the bases raised to the 1/3 and 1/2 (or then why if the bases (x2+4)^3/2, and (3x+5)^4/3 aren't less than the others why are they the terms that are factored out).

4. ## Re: Factoring Question

Originally Posted by KhanDisciple
Thanks for the reply, however I still don't understand how the bases raised to the 3/2, and 4/3 are less than the bases raised to the 1/3 and 1/2 (or then why if the bases (x2+4)^3/2, and (3x+5)^4/3 aren't less than the others why are they the terms that are factored out).
they're not ...

when you factor $(x^2+4)^{1/2}$ out from $(x^2+4)^{3/2}$ you get $(x^2+4)^{1/2}\left[(x^2+4)^1]$

check it by distributing the common factor

5. ## Re: Factoring Question

Thank you for clarifying skeeter, I see what I was doing wrong now.