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

1 Answer
George C.
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#.

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