First, let’s rewrite the inequality as an equation and then solve it:

[latex]x^2 = 3x+4[/latex]

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

[latex](x+1)(x-4)=0[/latex]

[latex]x+1=0[/latex] or [latex]x-4=0[/latex]

[latex]x=-1[/latex] or [latex]x=4[/latex]

Second, plot the solutions on the number line. Since the inequality has “[latex]\leq[/latex]” (less than or equal to), which has “[latex]=[/latex]” (equal sign), they should be closed circles:

Note that now there are three intervals on the number line: [latex](-\infty, -1], [-1, 4],[/latex] and [latex][4, \infty)[/latex]

Third, choose your choice of test value from each interval:

I chose [latex]-2, 0,[/latex] and [latex]5[/latex] from those three intervals.

Fourth, test each interval using the test value:

When [latex]x=-2[/latex], [latex](-2)^2 \leq 3(-2)+4 \rightarrow 4 \leq -2[/latex] (False)  [latex]\therefore[/latex] not a solution

When [latex]x=0[/latex], [latex](0)^2 \leq 3(0)+4 \rightarrow 0 \leq 4[/latex] (True)  [latex]\therefore[/latex] solution

When [latex]x=5[/latex], [latex](5)^2 \leq 3(5)+4 \rightarrow 25 \leq 19[/latex] (False)  [latex]\therefore[/latex] not a solution

We can see the results of test values are true from the graph of [latex]f(x)=x^2-3x-4[/latex] below:

Therefore, the solution of the polynomial inequality [latex]x^2 \leq 3x+4[/latex] is [latex][-1, 4][/latex].

First, let's rewrite the inequality as an equation and then solve it:

[latex]-\frac{1}{3}x^2 = 0[/latex]

[latex]x^2=0[/latex]

[latex]x=0[/latex]

Second, plot the solution on the number line. Since the inequality has "[latex]\geq[/latex]" (greater than or equal to), which has "[latex]=[/latex]" (equal sign), it should be closed circles:

Third, choose your choice of test value from the interval:

I chose [latex]-1[/latex] and [latex]1[/latex].

Fourth, test the interval using the test value:

When [latex]x=-1[/latex], [latex]-\frac{1}{3}(-1)^2 \leq 0 \rightarrow -\frac{1}{3} \leq 0[/latex] (True)  [latex]\therefore[/latex] solution

When [latex]x=1[/latex], [latex]-\frac{1}{3}(1)^2 \leq 0 \rightarrow -\frac{1}{3} \leq 0[/latex] (True)  [latex]\therefore[/latex] solution

We can see the result of test values are true from the graph of [latex]f(x)=-\frac{1}{3} x^2[/latex] below:

* The graph is below the [latex]x[/latex]-axis everywhere except at [latex]x=0[/latex]. However, since the inequality has "[latex]\leq[/latex]," which has "[latex]=[/latex]" (equal sign), [latex]x=0[/latex] also satisfies the inequality.

Therefore, the solution of the polynomial inequality [latex]-\frac{1}{3} x^2[/latex] is [latex](-\infty, \infty)[/latex].