Finding the Roots of a Polynomial

1. Finding the Roots of a Polynomial

2. Classifying the Roots of a Polynomial Describe the amount of roots and what number set they belong to for each graph: 2 Real Roots and 1 Repeated Real Root because it has three x-intercepts and one “bounces off” 1 Real Repeated Root and 2 Complex Roots because it only has 1 x-intercept that “bounces off” but has 4 “turns” 2 Real Roots and 2 Complex Roots because it only has 2 x-intercepts but has 4 “turns” 4 Complex Roots because it has no x-intercepts but has 4 “turns” 4 Real Roots because it has four x-intercepts Degree = 4 (there are 4 changes of directions) An nth degree Polynomial ALWAYS has n roots

3. Here is an Interesting Question.

4. Can an Odd-Degree Polynomial have Zero Real Roots? We have seen examples of even degree polynomials that have zero real roots. Can an odd degree polynomial have zero real roots? NO! Since odd degree polynomials have opposite end behavior and are continuous (no gaps), they must intersect the x-axis at least once.

5. Let’s try an example.

6. Example 1: Finding the Roots of a Polynomial Find the roots of y=x4 – 11x3 + 29x2+ 35x – 150. Use the graph or table of the cubic to find the roots. The three roots that can be found from the graph or table are x = -2, 3, and 5 because they are the x-intercepts. Since the equation is degree 4, it must have a total of 4 roots. But the graph “bounces off” the x-intercept 5. This means it is a double root. Thus all of the roots have been found. Therefore the roots are:

7. Let’s try another example.

8. Example 2: Finding the Roots of a Polynomial Find the roots of y=x4 – 3x3 – 2x2+ 4x. Use the graph or table of the degree 4 equation to find all 4 roots. The two EXACT roots that can be found from the graph or table are x = 0 and 1. The other two x-intercepts are irrational roots (not in a table). To find the irrational roots, we need to factor the polynomial. Notice x is a GCF. We can factor it out: Since x = 1 is a root, (x – 1) is a factor. Now use polynomial division to “factor out” the (x – 1) of the cubic.

9. Example 2: Finding the Roots of a Polynomial Find the roots of y=x4 – 3x3 – 2x2+ 4x. Current Factored form: Rewrite the polynomial: x2 -2x -4 x = 1 This quadratic can not be factored because the x-intercepts are irrational. x x3 -2x2 -4x x = 0 – 1 -x2 2x 4 Use the quadratic formula to find the roots. x3 – 3x2 – 2x + 4 In order to find the roots, find out when each factor is equal to 0. The first two factor’s zeros are the roots we found from the graph. The third factor’s zeros are the missing two irrational roots.

10. Example 2: Finding the Roots of a Polynomial Find the roots of y=x4 – 3x3 – 2x2+ 4x. Find the roots of x2 – 2x – 4 with the quadratic formula: Current Factored form: Therefore, the four roots are:

11. Let’s try another example.

12. Example 3: Finding the Roots of a Polynomial Find the roots of y=x3+ 6x2+ 11x+ 12. Use the graph or table of the cubic to find the roots. The only root that can be found from the graph or table is x = -4 because it is the only x-intercept. Since the equation is a cubic, it must have a total of 3 roots. Thus, this equation must have 2 complex roots. In order to find the two missing complex roots, we must factor the cubic equation. Since x = -4 is a root, (x + 4) is a factor of the cubic. Now use polynomial division to “factor out” the (x + 4) of the cubic.

13. Example 3: Finding the Roots of a Polynomial Find the roots of y=x3+ 6x2+ 11x+ 12. Rewrite the cubic: x2 2x 3 x x3 2x2 3x This quadratic has no real roots since it does not have x-intercepts. x = -4 + 4 4x2 8x 12 Use the quadratic formula to find the roots. x3 + 6x2 + 11x + 12 In order to find the roots, find out when each factor is equal to 0. The first factor’s zero is the root we found from the graph. The second factor’s zeros are the missing two complex roots.

14. Example 3: Finding the Roots of a Polynomial Find the roots of y=x3+ 6x2+ 11x+ 12. Find the roots of x2 + 2x + 3 with the quadratic formula: Therefore, the three roots are: