Polynomials

**Rocher** · December 13th 2007, 04:25 PM

Could you guys do these questions step by step? I have some difficulty understanding =/

$1) (3x^3+5x^4+4x^2+1)+(2x^3+3x-12x^4-1)$

$2) (y^3+7y+6y^2-6)-(7+6y+5y^3+4y^4)$

$3) (4-6x-x^3-x^4)-(2+3x^3+7x+5x^2)$

$4) (x+y)(x+2y)(2x-y)$

$5) (y^2+4)\div(y+1)$

$6) (2a^3-11a^2+7a-10)\div(a-5)$

$7) 4(b^3+5b+2+4b^2)\div(1+2b)$

$8) (x^3-2x^2-75)\div(x-5)$

$9) (6n^3+5n^2+10-13n)\div(3n-5)\$

$10) (2x^3-3x^2y-3xy^2-5y^3)\div(x^2+xy+y^2)$

Thank you very much!!

**TKHunny** · December 13th 2007, 06:22 PM

You have 50 posts and you haven't figured out that no one else wants to do your homework for you?! Truly, I am astounded.

Show your work. Really. Come on...

**Krizalid** · December 13th 2007, 06:29 PM

We actually don't do whole homework. Some of those problems are painful to explain. (Long division, for example.) So as TKHunny says, show your work or tell us where you're getting stuck.

**topsquark** · December 14th 2007, 08:56 AM

Originally Posted by Rocher

$1) (3x^3+5x^4+4x^2+1)+(2x^3+3x-12x^4-1)$

Combine like terms:
$(3x^3+5x^4+4x^2+1)+(2x^3+3x-12x^4-1)$

$= 3x^3+5x^4+4x^2+1+2x^3+3x-12x^4-1$

$= (3x^3+ 2x^3)+(5x^4 - 12x^4)+4x^2+(1 - 1)+3x$

$= 5x^3 - 7x^4+4x^2+3x$ <-- Now put it in standard order

$= -7x^4 + 5x^3 + 4x^2 + 3x$

Originally Posted by Rocher

$2) (y^3+7y+6y^2-6)-(7+6y+5y^3+4y^4)$

$3) (4-6x-x^3-x^4)-(2+3x^3+7x+5x^2)$

You try these now.

-Dan

**topsquark** · December 14th 2007, 09:00 AM

Originally Posted by Rocher

$4) (x+y)(x+2y)(2x-y)$

Tell you what, I'll do a similar problem.

$(3x - y)(x - 2y)(x + y)$

$= (3x - y)[(x - 2y)(x + y)]$

$= (3x - y)[x(x + y) - 2y(x + y)]$

$= (3x - y)[x^2 + xy - 2xy - 2y^2]$

$= (3x - y)[x^2 - xy - 2y^2]$

$= 3x[x^2 - xy - 2y^2] - y[x^2 - xy - 2y^2]$

$= 3x^3 - 3x^2y - 6xy^2 - x^2y + xy^2 + 2y^3$

$= 3x^3 - 4x^2y - 5xy^2 + 2y^3$

Can you do your problem now?

-Dan

**Rocher** · December 14th 2007, 09:13 PM

Originally Posted by topsquark

Tell you what, I'll do a similar problem.

$(3x - y)(x - 2y)(x + y)$

$= (3x - y)[(x - 2y)(x + y)]$

$= (3x - y)[x(x + y) - 2y(x + y)]$

$= (3x - y)[x^2 + xy - 2xy - 2y^2]$

$= (3x - y)[x^2 - xy - 2y^2]$

$= 3x[x^2 - xy - 2y^2] - y[x^2 - xy - 2y^2]$

$= 3x^3 - 3x^2y - 6xy^2 - x^2y + xy^2 + 2y^3$

$= 3x^3 - 4x^2y - 5xy^2 + 2y^3$

Can you do your problem now?

-Dan

Hmm, I got

$x^2+2xy-xy+xy+2y^2+2xy-y^2$
Is that right?

I just have one more question, it's really confusing me right now...

If x(4x+5) = Ax(2x+5)+Bx for all values of x, find the values of A and B.

My teacher taught me something like FOIL, which is first, outside, inner, last. So, I multiplied all of them together and I got $8x^2+20x+10x+25$ but I don't know where to go from there. Any help?

Also, there's also dividing polynomials that I just don't get. I understand addition, subtraction, multiplication now but division is just... so confusing.

Like for $\frac{4m+2}{2m-n+1} - \frac{2n}{n-2m-1}$ what in the world am I supposed to do!?! And why is it subtract sign? What does it mean? There are questions like $\frac{x^2+y^2}{a+b} \div \frac{7x+7y}{7a+7b}$ that have division sign, and all other signs. I realize I should show my steps, but I don't even know how to start these questions...

**badgerigar** · December 14th 2007, 09:56 PM

Is that right?

I don't think so. However I am not entirely sure which question you are attempting here but it doesn't look like the answer to any of your 10.

If x(4x+5) = Ax(2x+5)+Bx for all values of x, find the values of A and B.

You will need to expand out the brackets and then "equate coefficients". This means that you use the fact that the coefficients of $x^2$ , x and 1 must be the same on both sides of the equals sign. This should net you A and B

Like for

what in the world am I supposed to do!?! And why is it subtract sign?

Firs check whether you can cancel anything out in each fraction. Then, do what you would normally do with ordinary fractions to get them all over the same denominator. Then you can just subtract the numerators.

There are questions like

that have division sign, and all other signs. I realize I should show my steps, but I don't even know how to start these questions...

Turn the second fraction upside down and multiply.

**DivideBy0** · December 15th 2007, 12:05 AM

Originally Posted by Rocher

$\frac{4m+2}{2m-n+1} - \frac{2n}{n-2m-1}$

Hope you've attempted this yet,

$=\frac{4m+2}{2m-n+1}-\frac{2n}{n-2m-1}$

$=\frac{4m+2}{2m-n+1}-\frac{2n}{-(2m-n+1)}$

$=\frac{4m+2}{2m-n+1}+\frac{2n}{2m-n+1}$

$=\frac{4m+2n+2}{2m-n+1}$

**topsquark** · December 15th 2007, 07:03 AM

Originally Posted by Rocher

Hmm, I got

$x^2+2xy-xy+xy+2y^2+2xy-y^2$
Is that right?

First off, your answer is not simplified. Second, the answer has three factors of x and three factors of y in it. The result is a cubic in both x and y, not a quadratic as you posted.

Break this down into steps like I did. Multiply out the last two factors, then multiply the third into it.

-Dan

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