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Exercises Polynomials and factoring

Example 5 from Algebra and Trigonometry, 5.3 Graphs of Polynomial Functions

Finding the x-Intercepts of a Polynomial Function Using a Graph

Find the x-intercepts of $h (x) = x^{3} + 4 x^{2} + x - 6.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in Figure 6, it appears that there are x-intercepts at $x = −3, −2,$ and $1.$

Figure 6

We can check whether these are correct by substituting these values for $x$ and verifying that

$h (- 3) = h (- 2) = h (1) = 0$
Since $h (x) = x^{3} + 4 x^{2} + x - 6,$ we have:

$\begin{matrix} h (−3) & = & {(−3)}^{3} + 4 {(−3)}^{2} + (−3) - 6 = −27 + 36 - 3 - 6 = 0 \\ h (−2) & = & {(−2)}^{3} + 4 {(−2)}^{2} + (−2) - 6 = −8 + 16 - 2 - 6 = 0 \\ h (1) & = & {(1)}^{3} + 4 {(1)}^{2} + (1) - 6 = 1 + 4 + 1 - 6 = 0 \end{matrix}$
Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

$\begin{matrix} h (x) & = & x^{3} + 4 x^{2} + x - 6 \\ = & (x + 3) (x + 2) (x - 1) \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find the y- and x-intercepts of the function $f (x) = x^{4} - 19 x^{2} + 30 x .$

How did it go?

PreviousFinding the y- and x-Intercepts of a Polynomial in Factored Form Next Identifying Zeros and Their Multiplicities

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 5.3, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.