need one more zeros?

0 ⤊

need one more zeros?

find all zeros of the polynomial

p(x)=x^3-3x^2-9x-5

i found two of the 3 zeros and i got -1 and i know that is right but i seem cant to find the last zeros please find it for me

2007-10-11 12:47:23 · 3 answers · asked by star baller 360 5 in Science & Mathematics ➔ Mathematics

3 answers

The final 0 is 5
The three zeroes are -1, -1, 5

2007-10-11 12:59:58 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋

Divide (x^3 - 3x^2 - 9x - 5) by (x - (-1)), that is, by (x+1), using synthetic division if you have learned that, and you will get a quadratic that you should have no trouble factoring. As it turns out, -1 is a double root of p(x).

2007-10-11 19:58:56 · answer #2 · answered by Anonymous · 0⤊ 0⤋

consider the possibility of a double zero.

2007-10-11 20:07:47 · answer #3 · answered by davidosterberg1 6 · 0⤊ 0⤋