# Math Help - Helppp!!!! factor and solve

1. ## Helppp!!!! factor and solve

I have been sitting in my room for... 5 hours now, 5 hours!! Trying to do this algebra 2 sheet on quadratic equations and I am completley lost!!!

I can do the easy one's like
(t^2-9)=0
(t-3)(t+3)=0
t-3=0 t+3=0
t=3 t=-3

{3,-3}

the steps i was taught were
get zero on one side of equation
factor completely
set each to zero
solve

but I am stuck on these problems I can get them all set to zero and then that's it, i'm stuck after that
^2 = squared ^3 = cubed ^4=to the 4th power

1) t^3-t=0

2)3u^2+7u-6=0

3)x^2-6x-72=0

4)(s-1)^2(s-3)^2

5)0.2x^2-1.1x+0.5=0

6) (x^2-3)^2+(x^2-3)-2=0

7) (x^2-3x+2)^3=0

8) f(x)=(x+1)^4 +4(x+1)^3

If anyone could help me that would be amazing!!!!!!! I need help a.s.a.p!!!!

2. Originally Posted by mathallychallenged
I have been sitting in my room for... 5 hours now, 5 hours!!
If you were watching TV Simultaneously it does not count. But if that is true something to be proud of, most time I ever spend was 3 hours, trying to understand a proof to a theorem.

I do the harder ones.

7) (x^2-3x+2)^3=0
$(x^2-3x+2)(x^2-3x+2)(x^2-3x+2)=0$
$x^2-3x+2=0$
$(x-1)(x-2)=0$
8) f(x)=(x+1)^4 +4(x+1)^3
$(x+1)^3((x+1)+4)=0$
$(x+1)(x+1)(x+1)(x+5)=0$
$(x+1)=0$
$(x+5)=0$

3. Hello, mathallychallenged!

The steps i was taught were:
. . get zero on one side of equation
. . factor completely
. . set each to zero
. . solve . . . . . . . . Correct!

In some of these, the factoring may be tricky.

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

Then we have: . $\begin{Bmatrix}t \,= \,0 & \Rightarrow &t \,=\,0\\ t - 1 \:=\:0& \Rightarrow & t\,=\,1 \\ t + 1 \:=\:0 & \Rightarrow & t \,=\,\text{-}1\end{Bmatrix}$

$2)\;\;3u^2+7u-6\:=\:0$
Factor: . $(3u - 2)(u + 3)\:=\:0$

Then we have: . $\begin{Bmatrix}3u - 2 \:=\:0 & \Rightarrow & u \,=\,\frac{2}{3} \\ u + 3 \:=\:0 & \Rightarrow & u = \text{-}3\end{Bmatrix}$

$3)\;\;x^2-6x-72\:=\:0$
Factor: . $(x - 12)(x + 6)\:=\:0$ . . . etc.

$4)\;\;(s-1)^2(s-3)^2$
This is not an equation.
I assume it's supposed to be: . $(s-1)^2(s-3)^2\:=\:0$

So we have: . $\begin{Bmatrix}(s-1)^2\:=\:0 & \Rightarrow & s-1 \:=\:0 & \Rightarrow & s\,=\,1 \\
(s-3)^2\:=\:0 & \Rightarrow & s-3\:=\:0 & \Rightarrow & s = 3\end{Bmatrix}$

$5)\;\;0.2x^2-1.1x+0.5\:=\:0$
Multiply by 10: . $2x^2 - 11x + 5\:=\:0$

Factor: . $(x - 5)(2x - 1)\:=\:0$ . . . etc.

$6)\;\;(x^2-3)^2+(x^2-3)-2\:=\:0$

Replace $x^2 - 3$ with $u$.

The equation becomes: . $u^2 + u - 2 \:=\:0$
. . which factors: . $(u - 1)(u + 2)\:=\:0$
. . and we get: . $\begin{Bmatrix} u = 1\\u=\text{-}2\end{Bmatrix}$

Since $u \,=\,x^2 - 3$
. . we have: . $\begin{Bmatrix}x^2 - 3 \:=\:1 & \Rightarrow & x^2 = 4 & \Rightarrow & x = \pm 2 \\
x^3 - 3 \:=\:\text{-}2 & \Rightarrow & x^2 = 1 & \Rightarrow & x = \pm 1 \end{Bmatrix}$

$7) (x^2-3x+2)^3\:=\:0$

Take the cube root of both sides: . $x^2 - 3x + 2 \:=\:0$

Factor: . $(x-1)(x-2) \:=\:0$ . . . etc.

$8)\;\; f(x)\:=\:(x+1)^4 +4(x+1)^3$ . Find the zeros.

We have: . $(x+1)^4 + 4(x+1)^3\:=\:0$

Factor: . $(x+1)^3\,[(x+1) + 4] \:=\:0\quad\Rightarrow\quad(x+1)^3(x+5)\:=\:0$

And we have: . $\begin{Bmatrix}(x+1)^3\:=\:0 & \Rightarrow & x + 1 \:=\:0 & \Rightarrow & x = \text{-}1 \\
x + 5 \:=\:0 & & \Rightarrow & & x = \text{-}5\end{Bmatrix}$