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The Rational Zero Theorem
The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear
somewhere in the list.
The Rational Zero Theorem
If f ( x ) = anxn^ + an-1xn -1^ +…+ a 1 x + a 0 has integer coefficients and (where is reduced) is a rational zero, then p is a factor of the constant term a 0 and q is a factor of the leading coefficient an.
p q
p q
EXAMPLE : Using the Rational Zero Theorem
List all possible rational zeros of f ( x ) = 15 x^3 + 14 x^2 - 3 x – 2.
Solution The constant term is – 2 and the leading coefficient is 15.
1 2 1 2 1 2 3 3 5 5 15 15
Possible rational zeros Factors of the constant term,^2 Factors of the leading coefficient, 15 1, 2 1, 3, 5, 15 1, 2, , , , , ,
=^ -
Divide 1 and 2 by 1.
Divide 1 and 2 by 3.
Divide 1 and 2 by 5.
Divide 1 and 2 by 15.
There are 16 possible rational zeros. The actual solution set to f ( x ) = 15 x^3 + 14 x^2 - 3 x – 2 = 0 is {-1, -^1 / 3 ,^2 / 5 }, which contains 3 of the 16 possible solutions.
EXAMPLE: Solving a Polynomial Equation
Solve: x^4 - 6 x^2 - 8 x + 24 = 0.
Solution The graph of f ( x ) = x^4 - 6 x^2 - 8 x + 24 is shown the figure below. Because the x -intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation.
x-intercept: 2
The zero remainder indicates that 2 is a root of x^4 - 6 x^2 - 8 x + 24 = 0.
EXAMPLE: Solving a Polynomial Equation
Solve: x^4 - 6 x^2 - 8 x + 24 = 0.
Solution Now we can rewrite the given equation in factored form.
( x – 2)( x^3 + 2 x^2 - 2 x - 12) = 0 This is the result obtained from the synthetic division.
x – 2 = 0 or x^3 + 2 x^2 - 2 x - 12 = 0 Set each factor equal to zero.
x^4 - 6 x^2 + 8 x + 24 = 0 This is the given equation.
Now we must continue by factoring x^3 + 2x^2 - 2x - 12 = 0
EXAMPLE: Solving a Polynomial Equation
Solve: x^4 - 6 x^2 - 8 x + 24 = 0.
Solution Now we can solve the original equation as follows.
( x – 2)( x^3 + 2 x^2 - 2 x - 12) = 0 This was obtained from the first synthetic division.
x^4 - 6 x^2 + 8 x + 24 = 0 This is the given equation.
( x – 2)( x – 2)( x^2 + 4 x + 6) = 0 This was obtained from the second synthetic division.
x – 2 = 0 or x – 2 = 0 or x^2 + 4 x + 6 = 0 Set each factor equal to zero. x = 2 x = 2 x^2 + 4 x + 6 = 0 Solve.
EXAMPLE: Solving a Polynomial Equation
Solve: x^4 - 6 x^2 - 8 x + 24 = 0.
Solution We can use the quadratic formula to solve x^2 + 4 x + 6 = 0.
Let a = 1, b = 4, and c = 6.
=^ -^ ^ -
We use the quadratic formula because x^2 + 4 x + 6 = 0 cannot be factored.
x b^ b^ ac a
=^ -^ ^ -
= - 2 i 2^ Simplify.
(^4 8) Multiply and subtract under the radical. 2
=^ -^ ^ -
= -^ ^ i -^^8 =^ 4(2)( 1) -^^ =^2 i^2
The solution set of the original equation is {2, - 2 - ii 2,- 2 + ii 2 }.
