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
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
$\displaystyle 6c^2-72=11c$
$\displaystyle 6c^2 - 11c - 72 = 0$
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 $\displaystyle c = -\frac{8}{3}$.
$\displaystyle 4x^3-12x^2+8x=0$
$\displaystyle 4x(x^2 - 3x + 2) = 0$
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.
$\displaystyle 2n^3-30n^2+100n=0$
$\displaystyle 2n(n^2 - 15n + 50) = 0$
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