How do you find all the zeros of #f(x)=4x^3-4x^2-7x-2#?

1 Answer
Jul 10, 2016

Answer:

#f(x)# has zeros: #x=-1/2# with multiplicity #2# and #x=2#.

Explanation:

#f(x) = 4x^3-4x^2-7x-2#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-2# and #q# a divisor of the coefficient #4# of the leading term.

That means that the only possible rational zeros are:

#+-1/4, +-1/2, +-1, +-2#

We can quickly see that #f(1)# and #f(-1)# cannot be zero, since #7# is odd and all the other coefficients are even.

We find:

#f(2) = 32-16-14-2 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#4x^3-4x^2-7x-2 = (x-2)(4x^2+4x+1)#

#color(white)()#
To find the remaining zeros I could just try the other rational values we identified and factor by the next one we find, but there's a shortcut:

Do you see the pattern #4,4,1# of the coefficients of the remaining quadratic?

Did you know that #441 = 21^2# ?

As a result, we can spot that #4x^2+4x+1# is a perfect square trinomial:

#4x^2+4x+1 = (2x+1)^2#

This shows us that #4x^2+4x+1# has a repeated zero #x=-1/2#

graph{4x^3-4x^2-7x-2 [-4.877, 5.123, -2.4, 2.6]}