# Thread: HELP needed with3 algebra problems, factor

1. ## HELP needed with3 algebra problems, factor

the section iam doing is called "5-12 solving equations by factoring"
and i cant seem to get a reasonable answer to these problems

6c^2-72=11c

4x^3-12x^2+8x=0

2n^3-30n^2+100n=0

please show how you did it so i can learn

2. Originally Posted by elfen_lied
the section iam doing is called "5-12 solving equations by factoring"
and i cant seem to get a reasonable answer to these problems

6c^2-72=11c

4x^3-12x^2+8x=0

2n^3-30n^2+100n=0

please show how you did it so i can learn
6c^2-72=11c
$\displaystyle 6c^2 - 11c - 72 = (2c - 9)(3c + 8); c = 9/2, c = -8/3$

4x^3-12x^2+8x=0
$\displaystyle 4x(x^2 - 3x + 2) = 4x (x - 1)(x - 2); x=0, x=1, x=2$

2n^3-30n^2+100n=0
$\displaystyle 2n(n^2 - 15n + 50) = 2n(n-5)(n-10); n=0, n=5, n=10$

3. Originally Posted by elfen_lied
6c^2-72=11c
$\displaystyle 6c^2-72=11c$

$\displaystyle 6c^2 - 11c - 72 = 0$
We want a pair of factors of $\displaystyle 6 \cdot -72 = -432$ (the coefficient of the $\displaystyle c^2$ term times the coefficient of the constant term) that add to -11:
1, -432 --> -431
2, -216 --> -214
3, -144 --> -141
etc.

At some point in the list we get to:
16, -27 --> -11
So we split the middle term: $\displaystyle -11c = 16c - 27c$:
$\displaystyle 6c^2 - 11c - 72 = 0$

$\displaystyle 6c^2 + 16c - 27c - 72 = 0$

$\displaystyle 2c(3c + 8) - 9(3c + 8) = 0$

$\displaystyle (2c - 9)(3c + 8) = 0$ (You can verify that this is our factorization by multiplying it out.)

So either $\displaystyle 2c - 9 = 0$ or $\displaystyle 3c + 8 = 0$.

Thus $\displaystyle c = \frac{9}{2}$ or $\displaystyle c = -\frac{8}{3}$.

Originally Posted by elfen_lied
4x^3-12x^2+8x=0
$\displaystyle 4x^3-12x^2+8x=0$

$\displaystyle 4x(x^2 - 3x + 2) = 0$
You can apply the method above (split the -3x into -x - 2x and factor by grouping as in the above), but you can probably factor this by looking at it. Either way:

$\displaystyle 4x(x - 2)(x - 1) = 0$

So x = 0, 1, or 2.

Originally Posted by elfen_lied
2n^3-30n^2+100n=0
$\displaystyle 2n^3-30n^2+100n=0$

$\displaystyle 2n(n^2 - 15n + 50) = 0$

To factor the second term we need to find a pair of numbers that are factors of $\displaystyle 1 \cdot 50 = 50$ that add up to -15. I get -5 and -10:

$\displaystyle 2n(n^2 - 15n + 50) = 0$

$\displaystyle 2n(n^2 - 5n - 10n + 50) = 0$

$\displaystyle 2n(n(n - 5) - 10(n - 5)) = 0$

$\displaystyle 2n(n - 10)(n - 5) = 0$

Thus n = 0, 5, or 10.

-Dan

4. Thanks!