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 x = -3, -1, and 4. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 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

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.

As with all inequalities, we start by solving the equality [latex]6 - 5t - {t}^{2}= 0[/latex]. The first step is to write this in factored form. 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. We know the function can only change from positive to negative at these values, so these divide the inputs into 3 intervals.

We could choose a test value in each interval and evaluate the function [latex]f\left(t\right) = 6-5t-{t}^{2} = 0[/latex] at each test value to determine if the function is positive or negative in that interval

From our test values, we can determine this function is positive when -6 < t < 1, or in interval notation, [latex]\left[-6,1\right][/latex].

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

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.

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

On a number line this would look like:

From our test values, we can determine this function is negative when [latex]-3\leq x< -1[/latex] or [latex]-1<x \leq 4[/latex], or in interval notation: [latex][-3,-1) \cup (-1,4][/latex].  Notice that -1 is not included since it causes division by zero and is, thus, not in the domain of [latex]f(x)[/latex].

This time, the rational inequality is not in factored form, and there is not a zero on the right side. We need to subtract the 5 to the other side and get common denominators: [latex]\dfrac{2x^{2}+6x+9}{x+3}-5>0 \\ \dfrac{2x^{2}+6x+9}{x+3}-\dfrac{5(x+3)}{x+3}>0 \\ \dfrac{2x^{2}+6x+9-5(x+3)}{x+3}>0 \\ \dfrac{2x^{2}+x-6}{x+3}>0 \\ \dfrac{(x+2)(2x-3)}{x+3}>0[/latex]

We will now solve [latex]\dfrac{(x+2)(2x-3)}{x+3}=0[/latex] which has solutions at [latex]x=-2,\dfrac{3}{2},-3[/latex]. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals.
We could choose a test value in each interval and evaluate the function [latex]f\left(x\right) = \dfrac{(x+2)(2x-3)}{x+3}[/latex] at each test value to determine if the function is positive or negative in that interval

From our test values, we can determine this function is positive when [latex]-3<x<-2[/latex] or [latex]x>\dfrac{3}{2}[/latex], or in interval notation: [latex](-3,-2) \cup \left(\dfrac{3}{2},\infty \right)[/latex].