How do you find the zeroes of p(x)= x^3-x^2-10x-8?
1 Answer
Aug 11, 2018
Explanation:
Given:
p(x) = x^3-x^2-10x-8
By the ratonal zeros theorem, any rational zeros of
That means that the only possible rational zeros are:
+-1, +-2, +-4, +-8
Trying each in turn, we find:
p(color(blue)(-1)) = (color(blue)(-1))^3-(color(blue)(-1))^2-10(color(blue)(-1))-8
color(white)(p(-1)) = -1-1+10-8 = 0
So
x^3-x^2-10x-8 = (x+1)(x^2-2x-8)
We can factor the remaining quadratic by finding a pair of factors of
x^2-2x-8 = (x-4)(x+2)
So the other two zeros are