# Thread: Polynomials & more....Need help ASAP

1. ## Polynomials & more....Need help ASAP

Hi, I'm new to this forum. Currently taking math since it's a required pre-requisite. I must pass in order to move onto my college core classes. Since it's been over 10 years since I've done math (in high school) I am at a complete loss and feel like a REAL idiot when I'm in class.

Here at the problems I am stumped on:

(5x3+15x2-25x) ÷ 5x

(a9)-3 ÷ (a-4)7

-4w3-16w2+20w

b3-216

u3+125v3

2h2-h-3=0

2w(4w+1)=1

m3+2m2-3m=0

3x(2x+1)=18

2. ## Correction...

I just realized in my original post - the exponents aren't coming up. You can email me to get the correct equations??

devonca@gmail.com

3. Hello, devonca!

Welcome aboard!
I think I know what you meant by these problems.

I'll take baby-steps in these problems.
I hope this is an effective review for you.

Simplify: . $\frac{5x^3+15x^2-25x}{5x}$
Make three fractions: . $\frac{5x^3}{5x} + \frac{15x^2}{5x} - \frac{25x}{5x}$

. . and reduce: . $x^2 + 3x - 5$

Simplify: . $\frac{(a^9)^{-3}}{(a^{-4})^7}$
We have: . $\frac{a^{-27}}{a^{-28}} \;=\;a^{(-27) - (-28)} \;=\;a^1 \;=\;a$

Factor: . $-4w^3 - 16w^2 + 20w$
We have: . $-4w(w^2 + 4w - 5) \;=\;-4w(w-1)(w+5)$

Factor: . $b^3-216$ . . . . difference of cubes
We have: . $(b)^3 - (6)^3$

Therefore: . $(b - 6)(b^2 + 6b + 36)$

Factor: . $u^3+125v^3$ . . . . sum of cubes
We have: . $(u)^3 + (5v)^3$

Therefore: . $(u + 5v)(u^2 - 5uv + 25v^2)$

Solve: . $2h^2 - h - 3\:=\:0$
Factor: . $(h + 1)(2h - 3) \:=\:0$

Then: . $\begin{array}{ccccccc}h + 1 & = & 0 & \Rightarrow & h & = &\text{-}1 \\
2h - 3 & = & 0 & \Rightarrow & h & = & \frac{3}{2} \end{array}$

Solve: . $2w(4w+1)\:=\:1$
We have: . $8w^2 + 2w \:=\:1\quad\Rightarrow\quad 8w^2 + 2w - 1 \:=\:0$

. . which factors: . $(2w-1(4w+1) \:=\:0$

. . and has roots: . $\begin{array}{ccccccc}2w-1 & = & 0 & \Rightarrow & w & = & \frac{1}{2} \\
4w + 1 & = & 0 & \Rightarrow & w & = & \text{-}\frac{1}{4}\end{array}$

Solve: . $m^3 + 2m^2 - 3m\:=\:0$
Factor: . $m(m^2 + 2m - 3)\:=\:0\quad\Rightarrow\quad m(m-1)(m+3) \:=\:0$

And we have three roots: . $m \;=\;0,\:1,\:-3$

Solve: . $3x(2x+1)\:=\:18$

We have: . $6x^2 + 3x \:=\:18\quad\Rightarrow\quad 6x^2 + 3x - 18\:=\:0$

. . which factors: . $3(2x-3)(x+2) \:=\:0$

. . and has roots: . $x \;=\;\frac{3}{2},\:-2$