Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Do Now:

Aim: How do we solve polynomial equations using factoring?

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)

x3 + 3x2 + 3x2 + 9x + 2x + 6

Multiply by distribution of resulting factors

Combine like terms

x3 + 6x2 + 11x + 6

cubic expression is standard form

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Solving Polynomials by Factoring

Factor: 2x3 + 10x2 + 12x

GCF

Factor trinomial

2x(x2 + 5x + 6x)

2x(x + 3)(x + 2)

Solve: 2x3 + 10x2 + 12x = 0

2x(x2 + 5x + 6x) = 0

2x(x + 3)(x + 2) = 0

Set factors equal to zero(Zero Product Property)

2x = 0(x + 3) = 0(x + 2) = 0

x = 0, -2, -3

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Graphical Solutions

Solve: 2x3 + 10x2 + 12x = 0 2x(x + 3)(x + 2) = 0

x = 0, -2, -3

2

-2

-4

f x = 2x3+10x2+12x

ax3 + bx2 + cx + d = y

Cubic equation in Standard form

y-intercept

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

4

2

-2

-4

-6

-8

Model Problem

Find the zeros of y = (x – 2)(x + 1)(x + 3)

0 = (x – 2)(x + 1)(x + 3)

Set factors equal to zero(x – 2) = 0(x + 1) = 0(x + 3) = 0

Zeros/roots/x-intercepts are found at y = 0 (x-axis)

x = -3, -1, 2y-intercept?

(-2)(1)(3) = -6

4

2

-2

-4

-6

-8

is the product of last terms of binomial factors

y = x3 + 2x2 – 5x – 6

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Synthetic Division & Factors

Divide x3 – x2 + 2 by x + 1 using synthetic division

1 -1 0 2-1

1-1

-2 -2

0

x2 – 2x + 2 0quotient remainder

Since there is no remainder x + 1 is a factor of x3 – x2 + 2

Remainder Theorem

(x2 – 2x + 2)(x + 1) = x3 – x2 + 2

= 0x3 – x2 + 2 Solve: (x2 – 2x + 2)(x + 1) = 0

Set factors equal to zero(x2 – 2x + 2) = 0(x + 1) = 0

x = -1

Quadratic Formula

x = 1 ± i

22

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Synthetic Division, Factors and Graphing

= 0x3 – x2 + 2 Solve: (x2 – 2x + 2)(x + 1) = 0

Set factors equal to zero(x2 – 2x + 2) = 0(x + 1) = 0

x = -1

Quadratic Formula

x = 1 ± i

4

3

2

1

-1

-2

-2 2

q x = x3-x2 +2

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Model Problem

Use synthetic division to show that x + 3 is a factor of y = 2x3 + 11x2 + 18x + 9 then

solve 2x3 + 11x2 + 18x + 9 = 0

2 11 18 9-3

2-6

5 -9

0

2x2 + 5x + 3

-153 2x3 + 11x2 + 18x + 9 = 0

(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)

3

2

1

-1

-2

-4 -2

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Factor by Grouping

x3 – 2x2 – 3x + 6

Group terms (x3 – 2x2) – (3x – 6)

Factor Groups x2(x – 2) – 3(x – 2)

Distributive Property (x2 - 3)(x – 2)

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Model Problem

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)

Factor:

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Model Problem

x3 – 3x2 – 3x + 9 = 0

x2(x – 3) – 3(x – 3) = 0

Group

(x2 – 3)(x – 3) = 0

Factor

(x – 3) = 0 x = 3

(x2 – 3) = 0 x =

3

(x3 – 3x2) – (3x – 9) = 0

10

8

6

4

2

-2

3 1.73

Solve:

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Solving Quadratic Form Equations

x4 – 3x2 + 2 = 0

u2 – 3u + 2 = 0 Quadratic form

(x2 – 1)(x2 – 2) = 0

Factor

Substitute

(x – 1)(x + 1)(x2 – 2) = 0

(x – 1) = 0 x = 1

(x + 1) = 0 x = -1

(x2 + 2) = 0 x =

2

u = x2

u2 – 3u + 2 = 0

(u – 1)(u – 2) = 0

Factor

Set factors = 0& solve for x

Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig.

Model Problem

Solve: x4 – x2 = 12