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 Algebra Pre-Algebra and Basic Algebra Math Forum

 November 9th, 2008, 06:31 PM #1 Newbie   Joined: Nov 2008 Posts: 1 Thanks: 0 Can anyone help me with Polynomials? Hi i need some help with 4 math questions if anyone knows how to do them *with the whole process can you please solve it for me. I got a bet with my friend on like 10 bucks for this. If you do answer please answer with full process and i will best answer you. 1st question 3x^2(x-1)+3x^3(x+2) 2nd question 5/3 (1/3z-1) - 5/9z(1+1/2) 3rd question (4x-y)(1-y)-[(x+y^2)(2-x)] 4th question 2x(3x2-2)+(1/3x^2)^2(0.2x+3) the #/# is a fraction. "#" is any number. November 9th, 2008, 08:44 PM   #2
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Re: Can anyone help me with Polynomials?

I will do the first one. The process is not so difficult, and I suspect that you will learn it quickly.

Quote:
 1st question 3x^2(x-1)+3x^3(x+2)
We first distribute. so 3x^2(x - 1) becomes (3x^2 * x) - (3x^2 * 1) = 3x^3 - 3x^2
Similarly, we distribute the second one: 3x^3(x + 2) = 3x^3 * x + 3x^3 * 2 = 3x^4 + 6x^3

So in all we have 3x^3 - 3x^2 + 6x^3 + 3x^4 =
3x^4 + 7x^3 - 3x^2

Does that help? November 9th, 2008, 08:46 PM   #3
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Joined: Apr 2008

Posts: 435
Thanks: 0

Re: Can anyone help me with Polynomials?

I will do the first one. The process is not so difficult, and I suspect that you will learn it quickly.

Quote:
 1st question 3x^2(x-1)+3x^3(x+2)
We first distribute. so 3x^2(x - 1) becomes (3x^2 * x) - (3x^2 * 1) = 3x^3 - 3x^2
Similarly, we distribute the second one: 3x^3(x + 2) = 3x^3 * x + 3x^3 * 2 = 3x^4 + 6x^3

So in all we have 3x^3 - 3x^2 + 6x^3 + 3x^4 =
3x^4 + 9x^3 - 3x^2

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