How do you find the zeroes of p(x)= x^3-x^2-10x-8?

1 Answer
Aug 11, 2018

p(x) has zeros -1, 4 and -2

Explanation:

Given:

p(x) = x^3-x^2-10x-8

By the ratonal zeros theorem, any rational zeros of p(x) are expressible in the form p/q for integers p, q with p a divisor of the constant term -8 and q a divisor of the coefficient 1 f the leading term.

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=-1 is a zero and (x+1) a factor:

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 8 which differ by 2, namely 4 and 2, and hence:

x^2-2x-8 = (x-4)(x+2)

So the other two zeros are x=4 and x=-2.