Solving Polynomial Inequalities

This polynomial is already isolated on the left-hand side of the inequality, so that step is done.

Now, I factor the polynomial to find all of its zeroes:

Setting each factor equal to zero and solving, I get the polynomial's zeroes as being:

(Review how to solve polynomials, if you're not sure how to get this result.)

These zeroes split the number line into the following intervals:

(The intervals have square brackets at their numerical endpoints because this is an "or equal to" inequality.)

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To solve by the Test-Point Method, I would pick a sample point in each interval. As you can see, if your polynomial or rational function has many factors, the Test-Point Method can become quite time-consuming, multiplying the opportunities for error.

On the other hand, to solve by the Factor Method, I solve each factor for its positivity:

Then I draw my grid, where the rows are the factors, the vertical lines mark the zeroes that split the number line into intervals, the boxes formed by these lines will contain the signs of the factors on each interval, and the top row will contain the signs on the intervals for the entire polynomial:

For each factor, I put a "plus" sign in each box where that factor is greater than zero; then I put "minus" signs in all the rest of that row's boxes:

Now I apply the power of the Factor Method.

The polynomial is the product of all of its factors. On each interval, the sign of the polynomial is the product of the signs of its factors. And I can find the sign of the product by counting up "minus" signs. If there are an odd number of "minus" signs (that is, if there are an odd number of negative factors) on that interval, then the sign on the polynomial on that interval is "minus"; if there are an even number of "minus" signs, then they cancel each other out and the polynomial is "plus" on that interval. I put the sign of the product of the factor-signs up in that top row.

So, looking at the column for the first interval in the table above, (−∞, −5], I see that there are five "minus" signs. The product of five "minus" signs is "minus", so I put a "minus" at the top of that column (below). This tells me that the complete polynomial is negative on this interval.

In the second column, representing the interval [−5, −3], there are four "minus" signs (above), which multiply to "plus". So I put a "plus" sign at the top of this column (below), telling me that the polynomial is positive on this interval.

And so forth, for the other intervals. My result is shown below:

Checking back on the original inequality, I see that I'm supposed to find where the polynomial is greater than or equal to zero, so I'll be picking the intervals with a "plus" sign at the top of an interval's column, and I'll need to remember to include the endpoints.

So my answer is:

The factored form of the above polynomial is:

Now I need to find where each factor is greater than zero.

In the second inequality above, note that the "greater than" became a "less than". You should expect "backwards" factors like this to change your interval signs in the opposite way from the usual.

In the last line above, I said that x² + 3 is always positive. I can tell, because this factor is a square (so, non-negative) added to a positive number; this factor must always be positive. I will get no x-intercepts from this factor.

In the two lines right above the line with the factor containing the square, I've solved the same factor twice. This is becasue it is included twice in the polynomial's factorization, and I'll want to remember to include both copies of this factor in my chart.

Checking the original exercise, I see that they're wanting the intervals where the polynomial is positive, and that I'll need to include the endpoints. So my answer is: