Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Do Now: Aim: How do we solve polynomial...
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Transcript of Aim: Solving Polynomials by Factoring Course: Alg. 2 & Trig. Do Now: Aim: How do we solve polynomial...
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