# What is the solution set for this problem?Solve the polynomial inequality and graph the solution set on a real number. Express the interval notation. x^3+x^2+16x+16<0

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### 1 Answer

Solve `x^3+x^2+16x+16<0` :

(1) Set the expression equal to 0 and solve:

`x^3+x^2+16x+16=0`

`x^2(x+1)+16(x+1)=0` Factor 2 terms at a time

`(x^2+16)(x+1)=0` Distributive property

By the zero product property either `x^2+16=0` which has no solutions in the real numbers, or `x+1=0==>x=-1`

(2) The zeros of the function divide the real numbers into intervals, In this case x<-1,x>-1 or `(-oo,-1)uu(-1,oo)`

We examine the function on those intervals by picking a test number.

`f(-2)=(-2)^3+(-2)^2+16(-2)+16=-20` so every x in the interval x<-1 is a solution.

`f(0)=16` so no x in the interval x>-1 is a solution.

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**(3) The solution is x<-1 (Since this is a strict inequality we do not include the -1)**

**The graph of the solution:**

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The graph of the function:

Notice that for the solution set you can trace along the x-axis, and everywhere the function is negative you have a solution.

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