the problem is worked out, but i need to find out how. The book is not very helpful in explaining this.
numerator is x^4 - 8x
denominator is 3x^3 - 2x^2 - 8x
The answer is given as:
numerator: x^2 +2x +4
denominator: 3x +4
2007-02-02 18:27:05 · 3 answers · asked by rwwebber4 2 in Science & Mathematics ➔ Mathematics
I am not understanding howx^3-8 factors out to (x-2)(x^2+2x+4)
Please explain
2007-02-02 19:35:51 · update #1
(x^4 - 8x) / (3x^3 - 2x^2 - 8x) =
factor an x out of both numerator and denominator:
x(x^3 - 8) / (x(3x^2 - 2x - 8) =
Cancel the x's and factor the remaining terms:
(x - 2)(x^2 + 2x + 4) / ((x - 2)(3x + 4) =
Cancel the (x - 2) factors.
(x^2 + 2x + 4) / (3x + 4)
2007-02-02 19:06:33 · answer #1 · answered by Helmut 7 · 1⤊ 0⤋
Well, to start you can factor an x out of the denominator and numberator to get: (x^3-8)/(3x^2-2x-8) The top factors as (x-2)*(x^2+2x+4) and the denominator factors as (3x+4)*(x-2); since both the denominator and the numerator have a factor of (x-2), you can cancel this out, which leaves you with the answer that you already have.
2007-02-03 02:33:13 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋
i used to do that back in 5th grade... romania has high standards.... but i forgot... woah america turned me into a dumbass....
2007-02-03 03:08:25 · answer #3 · answered by [Unknown] 2 · 0⤊ 3⤋