How would you solve : x^2(3x-2/3)(5x+1/3)?

Question

The back of the book shows the answer to be 15x^4 - 7/3x^3 - 2/9x^2
But I need to know how to get that.
(I used ^ to represent exponent)

Answer 1

first, ignore the x^2 for now. We'll do that at the end.

(3x-2/3) * (5x+1/3) is a binomial multiplication problem. This indicates to use the FOIL method. The FOIL method is an acronym for the act of multiplying terms from the two terms above. F stands for "First", O stands for "Outer", I for "Inner," and L for "Last." Lastly, you add up the 4 answers, and come out with a final answer.

So, we first multiply the first terms of both binomials: 3x * 5x. That gives us 15x^2. Next, we multiply the outer terms: 3x * 1/3. That gives us x. Next, we multiply the inner terms: -2/3 * 5x. That gives us -(10/3)x. Lastly, multiply the last terms: -2/3 * 1/3. That gives us -2/9.

Now, add these four parts up: 15x^2 + x -10x/3 - 2/9 = 15x^2 -7/3x -2/9.

Now, remember we decided to leave the x^2 until last. So now we'll multiply the whole expression through by x^2.

(15x^2 -7/3x -2/9) * x^2 = 15x^4 - 7/3 x^3 - 2/9 x^2.

This is the same as the answer your textbook gives.

Hope this helped :)

Answer 2

1. First, you multiply x^2 with the first binomial. Answer should be (3x^3 - 2/3x^2).
2. Multiply (3x^3 - 2/3x^2) by (5x + 1/3). Answer should come out to be 15x^4 + x^3 - 10/3x^3 - 2/9x^2.
3. Combine like terms. Your final answer should be 15x^4 - 7/3x^3 - 2/9x^2.

Answer 3

I think you factor it... Like, distribute the x^2 to the 3x, -2/3, then to the 5x, and finally the 1/3rd. Add the like items, if you can. I'm not sure if that'll work, but I seem to remember doing that. Give it a shot!

Answer 4

now here's the funny part...unless you are going to be an engineer or chemist, in 5 yrs this will be meaningless....in 10 years, nobody will give a crap if this is on your resume, because everyone has PC's to do this for them in the workplace, in 15 yrs at your reunion nobody will care, and in 20 yrs, robots will do everything for us....so why stress over it, and move on to an easier problem