6.5 Theorems About Roots of Polynomial Equations Rational Roots

POLYNOMIALS and THEOREMS Theorems of Polynomial Equations There are 4 BIG Theorems to know about Polynomials Rational Root Theorem Irrational Root Theorem Imaginary Root Theorem Descartes Rule

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

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!

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

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 2 3 4 (Some factors of 24) 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!

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 ao

Example (RRT) ±3, ±1 ±1 ±12, ±6 , ±3 , ± 2 , ±1 ±4 ±1 , ±3 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???

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!

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

Let’s Try One 5x3-24x2+41x-20=0

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.

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

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

Step 6 – synthetic division 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 (1)³ + (1)² – 3(1) – 3 = -4 1 -3 (-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

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

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.

Let’s Try One 2x3 – x2 + 2x - 1

Using the Polynomial Theorems FACTOR 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

Using the Polynomial Theorems FACTOR 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

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

Using the Polynomial Theorems FACTOR 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

Complex ConjugateTheorem For a polynomial If a + bi is a root, Then a – bi is also a root Complex roots always come in pairs. Real values do not. CONJUGATE ___________________________ Complex pairs of form a + bi and a - bi

Example 1. For polynomial has roots 3 + 2i 2 3 – 2i Other roots ______ Degree of Polynomial ______ 2. For polynomial has roots -1, 0, - 3i, 1 + i 6 3i, 1 - i Other roots __________ Degree of Polynomial ______

Irrational Root Theorem For a polynomial If a + √b is a root, Then a - √b is also a root Irrationals always come in pairs. Real values do not. CONJUGATE ___________________________ Complex pairs of form a + √ b and a - √ b

Example (IRT) 1. For polynomial has roots 3 + √2 2 3 - √2 Other roots ______ Degree of Polynomial ______ 2. For polynomial has roots -1, 0, - √3, 1 + √5 6 √3 , 1 - √5 Other roots __________ Degree of Polynomial ______

Example (IRT) 1. For polynomial has roots 1 + √3 and -√11 1 - √3 √11 Other roots ______ _______ 4 Degree of Polynomial ______ Question: One of the roots of a polynomial is Can you be certain that is also a root? No. The Irrational Root Theorem does not apply unless you know that all the coefficients of a polynomial are rational. You would have to have as your root to make use of the IRT.

Write a polynomial given the roots 5 and √2 Another root is - √2 Put in factored form y = (x – 5)(x + √2 )(x – √2 )

Decide what to FOIL first y = (x – 5)(x + √2 )(x – √2 ) -2

FOIL or BOX to finish it up (x-5)(x² – 2) y = x³ – 2x – 5x² + 10 Standard Form y = x³ – 5x² – 2x + 10 x2 -2 x -5 X3 -2x -5x2 10

Write a polynomial given the roots -√5, √7 Other roots are √5 and -√7 Put in factored form y = (x – √5 )(x + √5)(x – √7)(x + √7) Decide what to FOIL first

y = (x – √5 )(x + √5)(x – √7)(x + √7) Foil or use a box method to multiply the binomials X -√5 X -√7 x √5 x √7 X2 -X √5 X2 -X √7 X √5 -5 X √7 -7 (x² – 5) (x² – 7)

FOIL or BOX to finish it up y = x4 – 7x² – 5x² + 35 Clean up -7 X4 -5x2 -7x2 35