# Math Help - polynomials

1. ## polynomials

1- divide using long division (9x^3-9x^2-x-12) / (3x-5)

2- simplify 3/x^2-9 + 4/x-3

3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor

4- simplify 4x^5y^-3 / 20x^-2y

*****
sorry if the math is messy, but I don't know the code to make it nice.

2. for #2, i got:

3(x-3)/(x+3)(x-3)(x-3) + 4(x+3)(x-3)/(x-3)(x+3)

3(x-3) + 4(x^2-9)/x^2-9(x^2-9)

is that wrong?

3. Originally Posted by Leah<3
1- divide using long division (9x^3-9x^2-x-12) / (3x-5)
i don't know how to type up the long division, so try it and come up with an answer and i'll tell you if its right.

2- simplify 3/x^2-9 + 4/x-3
you don't have to know the codes (though they are not that hard, i started learning them just today, and i'm doing pretty well i think) but you can use parenthesis to simplify. is this what you mean?

Simplify $\frac {3}{x^2 - 9} + \frac {4}{x - 3}$

3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor
think of other factors of 8 and plug them into the formula, if the result is zero, then x - the thing you plugged in is a factor.

4- simplify 4x^5y^-3 / 20x^-2y
again, use parenthesis

should it be, simplify $\frac {4x^5 y^{-3}}{20x^{-2}y}$ ?

4. Originally Posted by Jhevon
i don't know how to type up the long division, so try it and come up with an answer and i'll tell you if its right.
i'm not sure how to do it, that's why i asked...

Originally Posted by Jhevon
you don't have to know the codes (though they are not that hard, i started learning them just today, and i'm doing pretty well i think) but you can use parenthesis to simplify. is this what you mean?

Simplify $\frac {3}{x^2 - 9} + \frac {4}{x - 3}$
yes. i tried it and got $\frac {3(x-3) + 4(x^2-9)}{x^2-9(x^2-9)}$

Originally Posted by Jhevon

think of other factors of 8 and plug them into the formula, if the result is zero, then x - the thing you plugged in is a factor.
....huh?

Originally Posted by Jhevon

should it be, simplify $\frac {4x^5 y^{-3}}{20x^{-2}y}$ ?
yes. i started working on it but got thrown off because of the negatives.

5. the codes take up so much time!!

6. Originally Posted by Jhevon

think of other factors of 8 and plug them into the formula, if the result is zero, then x - the thing you plugged in is a factor.
hm.. x-2? although i did try it and got -2.

7. Originally Posted by Leah<3
1- divide using long division (9x^3-9x^2-x-12) / (3x-5)

2- simplify 3/x^2-9 + 4/x-3

3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor

4- simplify 4x^5y^-3 / 20x^-2y

...
Hello,

to #1:
Code:
(9x³ - 9x² - x - 12) ÷ (3x - 5) = 3x² + 2x + 3
(3x³ - 15x²)
-------------
6x² - x
6x² - 10x
---------
9x - 12
9x - 15
--------
3   ←  remainder
to #2:
$\frac{3}{x^2-9} + \frac{4}{ x-3}=\frac{3}{ (x+3)(x-3)}+\frac{4(x+3)}{(x+3)(x-3)} = \frac{4x+15}{x^2-9}$

to #3:

a) use long division
b) use synthetic division
Code:
      | 1   -5   2   8
| 0   4   -4   -8
------|----------------
x = 4 | 1   -1   -2   0
That means: $x^3 -5x^2 + 2x + 8 = (x-4)(x^2 - x - 2)$
The last factor can be factorized again so that you finally have:
$x^3 -5x^2 + 2x + 8 = (x-4)(x- 2)(x+1)$

to #4:
$\frac{4x^5 y^{-3}}{20x^{-2} y} = \frac{1}{5} x^{5-(-2)} y^{-3 - 1} = \frac{1}{5} x^7 y^{-4} = \frac{x^7}{5y^4}$

8. Originally Posted by Leah<3
1- divide using long division (9x^3-9x^2-x-12) / (3x-5)
It's going to be kind of hard to explain this one. i'll do the problem by hand and try to scan a copy and hopefully you'll be able to figure it out from that.

2- simplify 3/x^2-9 + 4/x-3
$\frac {3}{x^2 - 9} + \frac {4}{x - 3} = \frac {3}{(x + 3)(x - 3)} + \frac {4}{x - 3}$ ....the LCD is (x + 3)(x - 3)
....................... $= \frac {3 + 4(x + 3)}{(x + 3)(x - 3)}$
....................... $= \frac {4x + 15}{(x + 3)(x - 3)}$

3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor
By the rational roots theorem, the possible rational roots are given by the factors of 8 are: $\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$

start with $\pm 1$ and start plugging in each of the factors of 8 into the formula, if we get zero as the result, we have a root.

we realize that -1 works, since, $(-1)^3 - 5(-1)^2 + 2(-1) + 8 = 0$

since x = -1 is a root, it means x + 1 is a factor by the factor theorem

4- simplify 4x^5y^-3 / 20x^-2y
$\frac {4x^5y^{-3}}{20x^{-2}y} = \frac {4}{20} \cdot \frac {x^5}{x^{-2}} \cdot \frac {y^{-3}}{y}$
............. $= \frac {1}{5} \cdot x^{5 - (-2)} \cdot y^{-3 - 1}$
............. $= \frac {1}{5} \cdot x^{7} \cdot y^{-4}$
............. $= \frac {x^7}{5y^4}$

EDIT: Thanks a lot earboth. anyway, i guess it's my own fault for going around asking a lot of questions before doing the problem