 # Rational Root Theorem.

## Presentation on theme: "Rational Root Theorem."— Presentation transcript:

Rational Root Theorem

Rational Root Theorem:
Suppose that a polynomial equation with integral coefficients has the root , where h and k are relatively prime integers. Then h must be a factor of the constant term of the polynomial and k must be a factor of the coefficient of the highest degree term. (useful when solving higher degree polynomial equations)

How many solutions does the equation have?
1. 4 2. 3

a) State all possible rational zeros for g(x) = x3 + 2x2 - 3x + 5
Factors of the constant Example 1 Factors of the leading coefficient a) State all possible rational zeros for g(x) = x3 + 2x2 - 3x + 5 +1 +5,+1 Possible rational zeros: +5, +1 b) State all possible rational zeros for g(x) = 6x3 + 6x2 - 15x - 2 +2,+1 +6,+3,+2,+1 +2,+1 Possible rational zeros: +6,+3,+2,+1

Solve using the Rational Root Theorem:
4x2 + 3x – 1 = 0 (any rational root must have a numerator that is a factor of -1 and a denominator that is a factor of 4) factors of -1: ±1 factors of 4: ±1,2,4 possible rational roots: (now use synthetic division to find rational roots) (note: not all possible rational roots are roots!)

Ex : Solve using the Rational Root Theorem:
possible rational roots:

Ex : Solve using the Rational Root Theorem:
possible rational roots:

Ex : Solve using the Rational Root Theorem:
possible rational roots: To find other roots can use synthetic division using other possible roots on these coefficients. (or factor and solve the quadratic equation)

Find all zeros of the polynomial function.
3. How many answers: 4 Factors of last term: 1, 2, 4 Factors of first term: 1 Possible rational zeros: 1, 2, 4

Find all real zeros of the function.
3. Possible rational zeros: 1, 2, 4 1 1 –4 3 4 –4 1 –3 4 1 –3 4 (x – 1) ( ) = 0 x3 – 3x2 + 0x + 4

Find all real zeros of the function.
3. Possible rational zeros: 1, 2, 4 1 1 –4 3 4 –4 1 –3 4 –1 1 –3 4 –1 4 –4 1 –4 4 (x – 1) (x + 1) ( ) = 0 x2 – 4x + 4 x –2 (x – 1)(x + 1)(x – 2)(x – 2) = 0 x –2 1, 2 x =

Find all zeros of the polynomial function.
4. How many answers: 4 1, 2, 3, 6, 9, 18 Factors of last term: Factors of first term: 1 Possible rational zeros: 1, 2, 3, 6, 9, 18

Find all real zeros of the function.
4. Possible rational zeros: 1, 2, 3, 6, 9, 18 1 1 –1 7 –9 –18 1 7 –2 1 7 –2 –20

Find all real zeros of the function.
4. Possible rational zeros: 1, 2, 3, 6, 9, 18 –1 1 –1 7 –9 –18 –1 2 –9 18 1 –2 9 –18 (x + 1) ( ) ( ) ( ) = 0 x3 – 2x2 + 9x – 18 x2 (x – 2) +9 (x – 2) (x + 1) (x2 + 9) (x – 2) (x + 1)(x2 + 9)(x – 2) = 0 x =

Find the roots of x3 + 8x2 + 16x + 5 = 0

Ex : Determine the zeros:

The Fundamental Theorem of Algebra
If f(x) is a polynomial of degree n where n>0, then the equation f(x) = o has at least one solution in the set of complex numbers.

Complex Conjugates Theorem
If f (x) is a polynomial function with real coefficients, and a+bi is an imaginary zero of f(x), then a-bi is also a zero of f(x). (imaginary numbers always come in pairs)

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and given zeros. 5. 3, –2 f(x) = (x – 3)(x + 2) x2 + 2x – 3x – 6 f(x) = x2 – x – 6

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and given zeros. 6. i, –2i f(x) = (x – i)(x + i)(x + 2i)(x – 2i) (x2 – i2) (x2 – 4i2) (x2 + 1) (x2 + 4) x4 + 4x2 + x2 + 4 f(x) = x4 + 5x2 + 4

Some possibilities for zeros (roots)
Equation Real Zeros Imaginary Zeros Linear 1 Quadratic 2 Cubic 3 Quartic 4 Quintic 5

Find the zeros. f(x) = x 3 – x2 + 4x – 4 = 0 g(x) = x 4 + 2x 3 – 5x 2 – 4x + 6 = 0

Write a polynomial function of least degree with a leading coefficient of 1.
2, 3, -4 (x – 2)(x – 3)(x + 4) = 0 (x2 – 5x + 6)(x + 4) = 0 x 3 - x2 – 14x + 24 = 0

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and given zeros. 7. x3 – 2x – 4x2 + 8 f(x) = x3 – 4x2 – 2x + 8

Write a polynomial function of least degree with a leading coefficient of 1.
7, i (x – 7)(x – i)(x + i) = 0 (x – 7)(x 2 + 1) = 0 x 3 – 7 x 2 + x – 7 = 0

Descartes’ Rule of Signs
Let f(x) be a polynomial function with real coefficients. The number of positive real zeros of f(x) is equal to the number of sign changes of the coefficients of f(x) or is less than this by an even number. The number of negative real zeros of f (x) is equal to the number of sign changes of the coefficients of f(-x) or is less than this by an even number.

huh ? ? ?

Using Descartes’ Rule of Signs….
f(x) = x6-2x 5+3x 4 – 10x 3-6x 2 -8x -8 3 sign changes, so f has 3 or 1 positive real roots f(-x) = x6+2x 5+3x x 3-6x 2 +8x -8 3 sign changes, so f has 3 or 1 negative real roots

g(x) = -x5 – 5x4 + 7x3 – 4x2 – 8x + 9 Determine the possible number of positive real zeros, negative real zeros, and imaginary zeros of the function given above. g(x) = -x5 – 5x4 + 7x3 – 4x2 – 8x + 9 g(-x) = x5 – 5x4 - 7x3 – 4x2 + 8x + 9

Example Determine the possible number of positive real zeros, negative real zeros, and imaginary zeros of the function given above.

Use Descartes’ Rule of signs to find the number of possible real roots.