Rational Root Theorem

1. Rational Root Theorem

2. Consider the following . . . x3 – 5x2 – 2x + 24 = 0 This equation factors to: (x+2)(x-3)(x-4)= 0 The roots therefore are: -2, 3, 4

3. Take a closer look at the original equation and our roots: x3 – 5x2 – 2x + 24 = 0 The roots therefore are: -2, 3, 4 What do you notice? -2, 3, and 4 all go into the last term, 24!

4. Let’s look at another • 24x3 – 22x2 – 5x + 6 = 0 • This equation factors to: • (x+)(x-)(x-)= 0 • The roots therefore are: -1/2, 2/3, 3/4

5. Take a closer look at the original equation and our roots: • 24x3 – 22x2 – 5x + 6 = 0 • This equation factors to: (x+1)(x-2)(x-3)= 0 234 The roots therefore are: -1, 2, 3 2 3 4 What do you notice? The numerators 1, 2, and 3 all go into the last term, 6! The denominators (2, 3, and 4) all go into the first term, 24!

6. This leads us to the Rational Root Theorem For a polynomial, If p/q is a root of the polynomial, then p is a factor of an and q is a factor of

7. Example (RRT) 1. For polynomial Here p = -3 and q = 1 Factors of -3 Factors of 1 ±3, ±1 ±1  Or 3,-3, 1, -1 Possible roots are ___________________________________ 2. For polynomial Here p = 12 and q = 3 Factors of 12 Factors of 3 ±12, ±6 , ±3 , ± 2 , ±1±4 ±1 , ±3  Possible roots are ______________________________________________ Or ±12, ±4, ±6, ±2, ±3, ±1, ± 2/3, ±1/3, ±4/3 Wait a second . . . Where did all of these come from???

8. Let’s look at our solutions ±12, ±6 , ±3 , ± 2 , ±1, ±4 ±1 , ±3 Note that + 2 is listed twice; we only consider it as one answer Note that + 1 is listed twice; we only consider it as one answer Note that + 4 is listed twice; we only consider it as one answer That is where our 9 possible answers come from!

9. Let’s Try One Find the POSSIBLE roots of 5x3-24x2+41x-20=0

10. Let’s Try One 5x3-24x2+41x-20=0

11. That’s a lot of answers! • Obviously 5x3-24x2+41x-20=0 does not have all of those roots as answers. • Remember: these are only POSSIBLE roots. We take these roots and figure out what answers actually WORK.

12. Step 1 – find p and q p = -3 q = 1 • Step 2 – by RRT, the only rational root is of the form… • Factors of p Factors of q

13. Step 3 – factors • Factors of -3 = ±3, ±1 Factors of 1 = ± 1 • Step 4 – possible roots • -3, 3, 1, and -1

14. Step 5 – Test each root • Step 6 – synthetic division X X³ + X² – 3x – 3 -1 1 1 -3 -3 -3 3 1 -1 (-3)³ + (-3)² – 3(-3) – 3 = -12 (3)³ + (3)² – 3(3) – 3 = 24 3 -1 0 (1)³ + (1)² – 3(1) – 3 = -4 1 0 -3 0 (-1)³ + (-1)² – 3(-1) – 3 = 0 THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR WHAT ROOTS WILL MAKE THE EQUATION =0 1x² + 0x -3

15. Step 7 – Rewrite • x³ + x² - 3x - 3 = (x + 1)(x² – 3) • Step 8– factor more and solve • (x + 1)(x² – 3) • (x + 1)(x – √3)(x + √3) • Roots are -1, ± √3

16. Let’s Try One Find the roots of 2x3 – x2 + 2x - 1 Take this in parts. First find the possible roots. Then determine which root actually works.

17. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 Step 1 – find p and q p = -6 q = 1 • Step 2 – by RRT, the only rational root is of the form… • Factors of p Factors of q

18. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 • Step 3 – factors • Factors of -6 = ±1, ±2, ±3, ±6 Factors of 1 = ±1 • Step 4 – possible roots • -6, 6, -3, 3, -2, 2, 1, and -1

19. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 • Step 5 – Test each root • Step 6 – synthetic division X x³ – 5x² + 8x – 6 -450 78 0 -102 -2 -50 -2 -20 -6 6 3 -3 2 -2 1 -1 3 1 -5 8 -6 THIS IS YOUR ROOT 6 3 -6 1 -2 2 0 1x² + -2x + 2

20. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 • Step 7 – Rewrite • x³ – 5x² + 8x – 6 = (x - 3)(x² – 2x + 2) • Step 8– factor more and solve • (x - 3)(x² – 2x + 2) • Roots are 3, 1 ± i X= 3 Quadratic Formula