# solve the polynomial inequality and graph the solutionset ona real number line, express the solutionset in interval notation x^3-4x^2-16x+64<0

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### 1 Answer

To solve the inequality, we need to find the roots of the associated polynomial `f(x)=x^3-4x^2-16x+64.`

Factoring the polynomial, we get:

`f(x)=x^3-4x^2-16x+64`

`=x^2(x-4)-16(x-4)`

`=(x-4)(x^2-4)`

`=(x-4)(x-4)(x+4)`

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

This means that the zeros of `f(x)` are x=-4 and x=4.

The real number line is now divided into three regions: `(-infty,-4)` , `(-4,4)` and `(4,infty)` .

For the left-most region, we take a point such as x=-5 and test the polynomial to get `f(-5)=(-9)^2(-5+4)=-81<0`

so this region is a solution of the inequality.

The centre region can be tested using a point, such as x=0 to get `f(0)=64>0` so this region is not a solution to the inequality.

The right-most region can also be tested using a point such as x=5, to get `f(5)=(5-4)^2(5+4)=9>0`

so this region is also not a solution to the inequality.

**The solution to the inequality is `(-infty,-4)` . A graph of the number line with the solution is **