Algebra 3
Section 6-4 (6-5 and 6-6): Solving Polynomial Equations
Leaning Target: To solve polynomial equations
Standard: A2.6.B and A2.6.D
Common Core: A-APR
A. Do Now: Solve each quadratic equation:
1. (x – 2)(x + 4) = 0
2. x2 + 7x – 18 = 0
3. x2 + 17 = 0
4. 3x2 – 3x + 10 = 0
B. Many of the methods used to solve quadratic equations will be used to solve
polynomial equations.
1. Factoring
2. Solving a square root
3. Zero-Product Property
4. Quadratic Formula
5. And some new methods: Synthetic Division, Cubes
C. Solving Polynomial Equations:
1. x3 + 10x2 + 24x = 0
a. Factor out the x
b. x(x2 + 10x + 24) = 0
c. Factor the quadratic
d. x(x + 4)(x + 6) = 0
e. Use the zero product property
f. x = 0, -4, -6
2. 0 = x4 – 3x2 – 28
a. Factor the polynomial as a quadratic
b. 0 = (x2 – 7)(x2 + 4)
c. Set up the zero product property
d. x2 – 7 = 0, x2 + 4 = 0
e. Solve each part by taking square-roots
f. x2 = 7, x2 = -4
g. x = ±√7, x = ±2i
h. What do you notice about the solutions (zeros) of each polynomial
function and the degree of each polynomial?
D. The Fundamental Theorem of Algebra: The solution to a polynomial equation
(function) of degree n will have n-roots (zeroes)
E. Solve the following polynomial equations:
1. x3 – 13x2 + 30x = 0
2. x4 – 13x2 + 36 = 0
3. x4 + 3x3 – 8x2 = 0
F. Solving Cubes: x3 – 27 or x3 + 64
1. Sum of Cubes: x3 + a3 = (x + a)(x2 – ax + a2)
a. x3 + 64
b. (x + 4)(x2 – 4x + 16)
2. Difference of Cubes: x3 – a3 = (x – a)(x2 + ax + a2)
a. x3 – 27
b. (x – 3)(x2 + 3x + 9)
3. Solve x3 – 8 = 0
a. Factor first: (x – 2)(x2 + 2x + 4) = 0
b. Solve the first factor: x = 2
c. Solve the second factor using the quadratic formula:
d. x = -2 ± √(-2)2 – 4(2)(4)
2(1)
x = -1 ± i√7
x = 2 and -1 ± i√7
4. Solve x3 + 125 = 0
G. Using Synthetic Division (If one of the factors is known):
1. Solve x3 – 2x2 – 5x + 10 = 0, if (x – 2) is a factor of the polynomial.
a. Use synthetic division to find the quotient.
b. 2 │1 -2 -5 10
│___ 2__ 0_ -10__
1
0 -5 0
c. x2 – 5 = 0
d. x2 = 5
e. x = ±√5
f. x = 2, ±√5
2. Solve 3x3 + x2 – x + 1 = 0, if (x + 1) is a factor of the polynomial.
3. Solve x3 – 4x2 + 9x – 36 = 0, if (x – 4) is a factor of the polynomial.
H. Ending Task: Find the roots of each polynomial function
1. x4 + 2x2 – 15 = 0
2. x3 – 3x2 – 9x = 0
3. x3 – 5x2 + 5x – 4 = 0 ((x – 4) is a factor)
4. x4 – 3x2 – 4 = 0