Expert: Sherman D. - 10/18/2004

Question
Hi Sherman,

Can you please show me the opeartion to factor the following and solve for X?

3Xsquared + 10X - 8 = 0

I think I could do if it were 2Xsquared + 10X - 8 = 0, but not 3X squared since I forgot to do it...
Thanks in advance.

Ang-sui


Answer
3x^2 + 10x - 8 = 0

To do it one way, you could do it like this

Using the factors of 3 and -8 will give you like 8 different possibilities

(3x + 1)(x - 8) = -24 + 1 = -23
(3x - 1)(x + 8) = 24 - 1 = 23
(3x + 4)(x - 2) = -6 + 4 = -2
(3x - 4)(x + 2) = 6 - 4 = 2
(3x + 8)(x - 1) = -3 + 8 = 5
(3x - 8)(x + 1) = 3 - 8 = -5
(3x + 2)(x - 4) = -12 + 2 = -10
(3x - 2)(x + 4) = 12 - 2 = 10

So the only possible factor is (3x - 2)(x + 4)

if you set each one to equal 0, like this

3x - 2 =
3x = 2
x = (2/3)
or
x + 4 = 0
x = -4

you get

x = (2/3) or -4

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But i like using the quadratic formula (conclusion to completeing the square), since if done right, this will guarantee your "x" values. However this only works for powers up to x^2.

I like to use the quadratic formula(conclusion to completeing the square)

x = (-b ± sqrt(b^2 - 4ac))/2a
where
a = 3
b = 10
c = -8

so

x = (-(10) ± sqrt((10)^2 - 4(3)(-8)))/2(3)
x = (-10 ± sqrt(100 - 4(-24)))/6
x = (-10 ± sqrt(100 + 96))/6
x = (-10 ± sqrt(196))/6
x = (-10 ± 14)/6
x = (-24/6) or (4/6)
x = -4 or (2/3)

using the factor formula (ax - b)(cx - d)
whereas "a" and "c" are the denominators, and "b" and "d" are the numerators. Using that, you get

(3x - 2)(x + 4) as your factor.

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As for powers up to x^3, i can only tell you this

ax^3 + bx^2 + cx + d = 0

(ax^3 + bx^2) + (cx + d) = 0

yx^2(ax + b) + z(cx + d) = 0

"y" will depending on what "a" and "b" are, as in if they can be factored further, same goes for "z" for "c" and "d". Anyway, if "a" = "c" and "b" = "d" then you can factor it to (yx^2 + z)(ax + b) if (yx^2 + z) can be factored further then factor it.

Example Problem

x^3 + 2x^2 - 9x - 18
(x^3 + 2x^2) + (-9x - 18)
x^2(x + 2) - 9(x + 2)
(x^2 - 9)(x + 2)
(x - 3)(x + 3)(x + 2)

By the way, 2x^2 + 10x - 8 doesn't have a perfect factor, but if it were 2x^2 + 10x + 8 or 2x^2 - 10x + 8, then that would work.