1. Aim: How do we solve polynomial equations using factoring?
Do Now:
Write the expression (x + 1)(x + 2)(x + 3) as a polynomial in standard form.
(x + 1)(x + 2)(x + 3)
FOIL first two factors
(x2 + 3x + 2)(x + 3)
Multiply by distribution of resulting factors
x3 + 3x2 + 3x2 + 9x + 2x + 6
Combine like terms
x3 + 6x2 + 11x + 6
cubic expression is standard form

2. Solving Polynomials by Factoring
Factor: 2x3 + 10x2 + 12x
GCF
2x(x2 + 5x + 6x)
Factor trinomial
2x(x + 3)(x + 2)
Solve: 2x3 + 10x2 + 12x = 0
2x(x2 + 5x + 6x) = 0
2x(x + 3)(x + 2) = 0
2x = 0
(x + 3) = 0
(x + 2) = 0
Set factors equal to zero
(Zero Product Property)
x = 0, -2, -3

3. Cubic equation in Standard form
Graphical Solutions
Solve: 2x3 + 10x2 + 12x = 0
2x(x + 3)(x + 2) = 0
x = 0, -2, -3
Cubic equation in Standard form
ax3 + bx2 + cx + d = y
y-intercept

4. Find the zeros of y = (x – 2)(x + 1)(x + 3)
Model Problem
Find the zeros of y = (x – 2)(x + 1)(x + 3)
Zeros/roots/x-intercepts are found at y = 0
(x-axis)
0 = (x – 2)(x + 1)(x + 3)
(x – 2) = 0
(x + 1) = 0
(x + 3) = 0
Set factors equal to zero
x = -3, -1, 2
y-intercept?
(-2)
(1)
(3)
= -6
is the product of last terms of binomial factors
y = x3 + 2x2 – 5x – 6

5. Synthetic Division & Factors
Divide x3 – x by x + 1
using synthetic division -1
Since there is no remainder x + 1 is a factor of x3 – x2 + 2 -1 2 -2 1 -2 2
x2 – 2x + 2
quotient remainder
Remainder Theorem
(x2 – 2x + 2)(x + 1) = x3 – x2 + 2
Solve:
x3 – x2 + 2
= 0
(x2 – 2x + 2)(x + 1) = 0
(x2 – 2x + 2) = 0
(x + 1) = 0
Set factors equal to zero
Quadratic Formula
x = -1
x = 1 ± i

6. Synthetic Division, Factors and Graphing
Solve:
x3 – x2 + 2
= 0
(x2 – 2x + 2)(x + 1) = 0
(x2 – 2x + 2) = 0
(x + 1) = 0
Set factors equal to zero
Quadratic Formula
x = -1
x = 1 ± i

7. Model Problem solve 2x3 + 11x2 + 18x + 9 = 0
Use synthetic division to show that x + 3 is a factor of y = 2x3 + 11x2 + 18x + 9 then
solve 2x3 + 11x2 + 18x + 9 = 0 -3 -6
-15 -9 2 5 3
2x3 + 11x2 + 18x + 9 = 0
2x2 + 5x + 3
(2x2 + 5x + 3)(x + 3) = 0
(2x + 3)(x + 1)(x + 3) = 0
(2x + 3) = 0
(x + 1) = 0
(x + 3) = 0
x = (-3, -3/2, -1) 7

8. Factor by Grouping x3 – 2x2 – 3x + 6 (x3 – 2x2) – (3x – 6)
Group terms
(x3 – 2x2) – (3x – 6)
Factor Groups
x2(x – 2) – 3(x – 2)
Distributive Property
(x2 - 3)(x – 2)

9. Model Problem Factor: 4x3 – 6x2 + 10x – 15 (4x3 – 6x2) + (10x – 15)
Group terms
(4x3 – 6x2) + (10x – 15)
Factor Groups
2x2(2x – 3) + 5(2x – 3)
Distributive Property
(2x2 + 5)(2x – 3)
Factor:
x3 – 2x2 – 4x + 8
(x3 – 2x2) – (4x – 8)
x2(x – 2) + 4(x – 2)
(x2 + 4)(x – 2) 9

10. Model Problem Solve: x3 – 3x2 – 3x + 9 = 0 (x3 – 3x2) – (3x – 9) = 0
Group
x2(x – 3) – 3(x – 3) = 0
Factor
(x2 – 3)(x – 3) = 0
(x – 3) = 0  x = 3
(x2 – 3) = 0  x =

11. Solving Quadratic Form Equations
x4 – 3x2 + 2 = 0
u = x2
u2 – 3u + 2 = 0
u2 – 3u + 2 = 0
Quadratic form
(u – 1)(u – 2) = 0
Factor
(x2 – 1)(x2 – 2) = 0
Substitute
(x – 1)(x + 1)(x2 – 2) = 0
Factor
(x – 1) = 0  x = 1
Set factors = 0
& solve for x
(x + 1) = 0  x = -1
(x2 + 2) = 0  x =

12. Model Problem
Solve: x4 – x2 = 12