3.5B Finding Real Roots of Polynomial Equations
Objectives:
A.APR.3: Identify zeros of polynomials when suitable factorizations are available, and use the
zeros to construct a rough graph of the function defined by the polynomial.
A.REI.11: Explain why the x-coordinates of the points where the graphs of the equations y = f(x)
and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions
approximately, … include cases where f(x) and/or g(x) are linear, polynomial, rational,
… functions.
For the board: You will be able to identify the multiplicity of roots.
You will be able to use the Rational Root Theorem and the Irrational Root Theorem to
solve polynomial equations.
Not all roots of a polynomial are integers but if a polynomial has rational roots, it is possible to find them.
Rational Root Theorem
If the polynomial P(x) has integer coefficients, then every rational root of the polynomial
equation P(x) = 0 can be written in the form p/q, where p is a factor of the constant term
of p(x) and q is a factor of the leading coefficient of P(x).
Open the book to page 184 and read example 3.
Example: Identify all the possible rational roots of the given equation, then narrow the search using the
graphing calculator, then check your solutions using synthetic division. Remember that once
you get a quadratic equation you have solving methods: factoring, square rooting, completing
the square, and the quadratic formula.
3x3 + 7x2 – 4 = 0
q: factors of 3: ±1, ±3
p: factors of 8: ±1, ±2, ±4
p/q: possible rational roots: ±1, ±2, ±4, ±1/3, ±2/3, ±4/3
Use a graph to narrow the search: -1, -2, 2/3
Use synthetic division/substitution to check.
1| 3 7 0  4
(x + 1)(3x2 + 4x – 4) = 0
3 4
3
4
4
4 |0
(x + 1)(3x – 2)(x + 2) = 0
x = -1, 2/3, -2
White Board Activity:
Practice: Identify all the possible rational roots of the equation, than narrow the search using the
graphing calculator, then check your solutions using synthetic division. Solve any resulting
quadratics by any of the methods previously discussed.
2x3 – 7x2 – 3x + 18 = 0
Factors of 2: ±1, ±2
Factors of 18: ±1, ±2, ±3, ±6, ±9, ±18
Possible rational roots: ±1, ±2, ±3, ±6, ±9, ±18, ±1/2, ±3/2, ±9/2
Narrow the search: 2, 3, -3/2
2 | 2  7 - 3  18
2
4
- 6  18
-3
-9 | 0
(x – 2)(2x2 – 3x – 9) = 0
(x – 2)(2x + 3)(x – 3) = 0
x = 2, -3/2, 3
This method also works for polynomial equations with integer solutions.
Example: Solve x3 + 3x2 – 4x – 12 = 0 and use your solutions to factor the polynomial.
Factors of 1: ±
1
Factors of 12: ±1, ±2, ±3, ±4, ±6
Possible rational zeros: ±1, ±2, ±3, ±4,
Graph y = x3 + 3x2 – 4x – 12 and narrow the search.
x = -3, -2, and 2
Check your guesses using synthetic division.
 3 |1 3
 4  12
(x + 3)(x2 – 4)= 0
3 0
12
(x + 3)(x – 2)(x + 2) = 0
x = -3, 2, -2
1 0
4 |0
White Board Activity:
Practice: Solve x3 + x2 – 6x - 18 = 0 and use your solutions to factor the polynomial.
Factors of 1: ±1
Factors of 6: ±1, ±2, ±3, ±6
Possible rational zeros: ±1, ±2, ±3, ±6,
Graph y = x3 - 6x2 + 11x – 6 = 0 and narrow the search.
x = 1, 2, 3
Check your guesses using synthetic division.
1 | 1 - 6 11  6
(x – 1)(x2 – 5x +6) = 0
1 -5
6
(x – 1)(x – 2)(x – 3) = 0
x = 1, 2, 3
1 -5
6 |0
Polynomial equations can have irrational roots.
Recall: If an equation can be reduced to a quadratic the irrational roots can be determined using the
quadratic formula.
Irrational Root Theorem
If the polynomial P(x) has rational coefficients and a + b c is a root of the polynomial
equation P(x) = 0, where a and b are rational and c is irrational, then a – b c is also
a root of P(x) = 0. (Irrational roots always appear as conjugate pairs.)
Recall: Conjugate pairs are
5 and - 5 or 1 + 5 and 1 - 5 .
Open the book to page 185 and read example 4.
Example: Identify all the real roots of 2x3 – 9x2 + 2 = 0.
Factors of 2: ±1, ±2
Possible rational roots: ±1, ±2, ±½
Narrowing: ½ , - ½
Checking ½ :
1/ 2 | 2
2
9
0
2
1
4
2
8
 4 |0
½)(2x2
(x –
– 8x – 4) = 0
2(x – ½)(x2 – 4x – 2) = 0
This resulting polynomial cannot be solved using factoring but can be solved using the
quadratic formula.
 (4)  (4) 2  4(1)(2) 4  16  8 4  2 6
x=


 2 6
2
2
2
x = ½, 2  6 , 2  6 (Note the conjugate pair.)
White Board Activity:
Practice:
a. Identify all of the real roots of 2x3 – 3x2 – 10x – 4 = 0
Factors of 2: ±1, ±2
Factors of 4: ±1, ±2, ±4
Possible rational x’s: ±1, ±2, ±4, ±½
Narrowing: -1, - ½
Checking -½ :
 1/ 2 | 2  3
 10  4
1
2
4
2
4
8
|0
(x + ½ )(2x2 – 4x – 8) = 0
2(x + ½ )(x2 – 2x – 4) = 0
 (2)  (2) 2  4(1)(4) 2  4  16 2  2 5
x=


 1 5
2
2
2
x = ½, 1 5 , 1 5
b. Identify all the real roots of x4 – 4x2 + 3 = 0
Factors of 1: ±1
Factors of 3: ±1, ±3
Possible rational x’s: ±1, ±3
Narrowing: 1, -1
Checking -1 :
1| 1
1| 1 0
4 0
3
1
1 3 3
1
1
3 3
|0
(x + 1 )(x3 – x2 – 3x + 3) = 0
(x + 1 )(x – 1)(x2 – 3) = 0
x2 – 3 =0
x2 = 3
x = 1, -1, ± 3
1
x=± 3
1
1
0
3
3
0 3
3
|0
Assessment:
Question student pairs.
Independent Practice:
Text: pg. 186 – 187 prob. 11 – 14, 24 – 26, 29 -34.
For a Grade:
Text: pg. 186 – 187 prob. 12, 24, 32.