# Aim: How do we solve polynomial equations using factoring? - PowerPoint PPT Presentation

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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. ( x 2 + 3 x + 2)( x + 3) . Multiply by distribution of resulting factors.

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Aim: How do we solve polynomial equations using factoring?

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### 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

### 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

### 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

### 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

-1

### Synthetic Division & Factors

Divide x3 – x2 + 2 by x + 1

using synthetic division

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

1 -1 0 2

2

-1

-2

0

1

2

-2

x2 – 2x + 2

0

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

x = -1

x = 1 ± i

### 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

x = -1

x = 1 ± i

### Model Problem

-3

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

-15

-6

-9

0

2

3

5

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)

### Factor by Grouping

x3 – 2x2 – 3x + 6

(x3 – 2x2) – (3x – 6)

Group terms

Factor Groups

x2(x – 2) – 3(x – 2)

Distributive Property

(x2 - 3)(x – 2)

### Model Problem

Factor:

4x3 – 6x2 + 10x – 15

(4x3 – 6x2) + (10x – 15)

Group terms

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)

(x2 – 3) = 0  x =

### Model Problem

Solve:

x3 – 3x2 – 3x + 9 = 0

Group

(x3 – 3x2) – (3x – 9) = 0

x2(x – 3) – 3(x – 3) = 0

Factor

(x2 – 3)(x – 3) = 0

(x– 3) = 0  x = 3

u2 – 3u + 2 = 0

x4 – 3x2 + 2 = 0

u = x2

u2 – 3u + 2 = 0

(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 =

### Model Problem

Solve: x4 – x2 = 12