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 1)Rational Root Theorem 2)Irrational Root Theorem 3)Imaginary Root Theorem 4)Descartes Rule

Consider the following... x 3 – 5x 2 – 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: x 3 – 5x 2 – 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 24x 3 – 22x 2 – 5x + 6 = 0 This equation factors to: (x+1)(x-2)(x-3)= The roots therefore are: -1/2, 2/3, 3/4

Take a closer look at the original equation and our roots: 24x 3 – 22x 2 – 5x + 6 = 0 This equation factors to: (x+1)(x-2)(x-3)= The roots therefore are: -1, 2, 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 a n and q is a factor of a o

1. For polynomial Possible roots are ___________________________________ Here p = -3 and q = 1 Factors of -3 Factors of 1  ±3, ±1 ±1 2. For polynomial Possible roots are ______________________________________________ Here p = 12 and q = 3 Factors of 12 Factors of 3  ±12, ±6, ±3, ± 2, ±1 ±4 ±1, ±3 Or ±12, ±4, ±6, ±2, ±3, ±1, ± 2/3, ±1/3, ±4/3 Or 3,-3, 1, -1 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 That is where our 9 possible answers come from! Note that + 4 is listed twice; we only consider it as one answer

Let’s Try One Find the POSSIBLE roots of 5x 3 -24x 2 +41x-20=0

Let’s Try One 5x 3 -24x 2 +41x-20=0

That’s a lot of answers! Obviously 5x 3 -24x 2 +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 5 – Test each root Step 6 – synthetic division X X³ + X² – 3x – (-3)³ + (-3)² – 3(-3) – 3 = -12 (3)³ + (3)² – 3(3) – 3 = 24 (1)³ + (1)² – 3(1) – 3 = -4 (-1)³ + (-1)² – 3(-1) – 3 = 0 THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR WHAT ROOTS WILL MAKE THE EQUATION = x² + 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 2x 3 – x 2 + 2x - 1 Take this in parts. First find the possible roots. Then determine which root actually works.

Let’s Try One 2x 3 – x 2 + 2x - 1

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

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

Step 5 – Test each root Step 6 – synthetic division X x³ – 5x² + 8x – THIS IS YOUR ROOT x² + -2x

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 Quadratic FormulaX= 3

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

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

1. For polynomial has roots 1 + √3 and -√11 Other roots ______ _______ 1 - √3 Degree of Polynomial ______ 4 √11 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 ) X -√2 x √2 X2X2 -X √2 X √2 -2 (x² – 2)

FOIL or BOX to finish it up (x-5)(x² – 2) y = x³ – 2x – 5x² + 10 Standard Form y = x³ – 5x² – 2x + 10 x 2 -2 x -5 X3X3 -2x -5x 2 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 -√7 x √7 X2X2 -X √7 X √7 -7 (x² – 7) X -√5 x √5 X2X2 -X √5 X √5 -5 (x² – 5)

y = (x² – 5)(x² – 7) FOIL or BOX to finish it up y = x 4 – 7x² – 5x² + 35 Clean up y = x 4 – 12x² + 35 x 2 -5 x 2 -7 X4X4 -5x 2 -7x 2 35