Question #9ed95

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Answer 1 · 2018-02-07T08:47:38

#x=4#

By inspection you can see that #(4)^3-(4)^2=64-16=48#.

If we rearrange we get #x^3-x^2-48=0#.

Knowing that #x=4# is a zero means that #x-4# is a factor.

Dividing we have:

#(x^3-x^2-48)/(x-4) = x^2+3x+12#

So we can say that:

#x^3-x^2-48 =(x-4)(x^2+3x+12)#

so

#x^3-x^2-48=0#

is equivalent to

#(x-4)(x^2+3x+12)=0#

We already known #x=4# is a solution.

#x^2+3x+12=0# has no real solutions since the discriminant is less than 0: #9-4(12)<0#.

The only solution is #x=4#, which I found by inspection, but we could also find by the rational root theorem.

Tony B