As with all inequalities, we start by solving the equality [latex]\left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)= 0[/latex], which has solutions at [latex]x = -3, -1[/latex], and [latex]4[/latex]. We know the function can only change from positive to negative at these values, so these divide the inputs into four intervals.
We could choose a test value in each interval and evaluate the function [latex]f\left(x\right) = \left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)[/latex] at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	f(test value)	> 0 or < 0
x < -3	-4	72	> 0
-3 < x < -1	-2	-6	< 0
-1 <  x < 4	0	-12	< 0
x > 4	5	288	> 0

On a number line this would look like:

From our test values, we can determine this function is positive when x < -3 or x > 4, or in interval notation, [latex]\left(-\infty, -3\right)\cup\left(4,\infty\right)[/latex]. We could have also determined on which intervals the function was positive by sketching a graph of the function. We illustrate that technique in the next example.

A square root is only defined when the quantity we are taking the square root of, the quantity inside the square root, is zero or greater. Thus, the domain of this function will be when [latex]6 - 5t - {t}^{2}\ge 0[/latex]. Again we start by solving the equality [latex]6 - 5t - {t}^{2}= 0[/latex]. While we could use the quadratic formula, this equation factors nicely to [latex]\left(6 + t\right)\left(1-t\right)=0[/latex], giving horizontal intercepts
t = 1 and t = -6.
Sketching a graph of this quadratic will allow us to determine when it is positive.
From the graph we can see this function is positive for inputs between the intercepts. So [latex]6 - 5t - {t}^{2}\ge 0[/latex] is positive for [latex]-6 \le t\le 1[/latex], and this will be the domain of the v(t) function.

In our other examples, we were given polynomials that were already in factored form, here we have an additional step to finding the intervals on which solutions to the given inequality lie. Again, we will start by solving the equality [latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} = 0[/latex]

Notice that there is a common factor of [latex]{x}^{2}[/latex] in each term of this polynomial. We can use factoring to simplify in the following way:

[latex]\begin{align}{x}^{4} - 2{x}^{3} - 3{x}^{2} &= 0&\\{x}^{2}\left({x}^{2} - 2{x} - 3\right) &= 0\\ {x}^{2}\left(x - 3\right)\left(x + 1 \right)&= 0\end{align}[/latex]

Now we can set each factor equal to zero to find the solution to the equality.

[latex]\begin{array}{ccc} {x}^{2} = 0 & \left(x - 3\right) = 0 &\left(x+1\right) = 0\\ {x} = 0 & x = 3 & x = -1\\ \end{array}[/latex].

Note that x = 0 has multiplicity of two, but since our inequality is strictly greater than, we don’t need to include it in our solutions.
We can choose a test value in each interval and evaluate the function

[latex]{x}^{4} - 2{x}^{3} - 3{x}^{2} = 0[/latex]

at each test value to determine if the function is positive or negative in that interval

Interval	Test x in interval	> 0,  < 0
x < -1	-2	x > 0
-1 < x < 0	-1/2	x <  0
0 < x < 3	1	x < 0
x > 3	5	x > 0

We want to have the set of x values that will give us the intervals where the polynomial is greater than zero. Our answer will be [latex]\left(-\infty, -1\right]\cup\left[3,\infty\right)[/latex].

 

The graph of the function gives us additional confirmation of our solution.