5 7applications of factoring
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Applications of Factoring
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Transcript of 5 7applications of factoring
Applications of Factoring
Applications of FactoringThere are many applications of the factored forms of polynomials.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3= 49 – 14 – 3
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3= 49 – 14 – 3 = 32
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3= 49 – 14 – 3 = 32
x2 – 2x – 3 = (x – 3)(x+1)
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3= 49 – 14 – 3 = 32
x2 – 2x – 3 = (x – 3)(x+1)We get (7 – 3)(7 + 1)
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3= 49 – 14 – 3 = 32
x2 – 2x – 3 = (x – 3)(x+1)We get (7 – 3)(7 + 1) = 4(8) = 32
Often it is easier to evaluate polynomials in the factored form.
Applications of FactoringThere are many applications of the factored forms of polynomials. Following are some of them:1. to evaluate polynomials,2. to determine the signs of the outputs,3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
We get 72 – 2(7) – 3= 49 – 14 – 3 = 32
x2 – 2x – 3 = (x – 3)(x+1)We get (7 – 3)(7 + 1) = 4(8) = 32
Often it is easier to evaluate polynomials in the factored form.
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2]
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4]
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2]
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3]
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2]
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Your turn: Double check these answers via the expanded form.
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Determine the signs of the outputs
Your turn: Double check these answers via the expanded form.
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Determine the signs of the outputsOften we only want to know the sign of the output, i.e. whether the output is positive or negative.
Your turn: Double check these answers via the expanded form.
Example B. Evaluate 2x3 – 5x2 + 2x for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2) = x(2x – 1)(x – 2)
For x = -2:(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9 For x = 3:3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Determine the signs of the outputsOften we only want to know the sign of the output, i.e. whether the output is positive or negative. It is easy to do this using the factored form.
Your turn: Double check these answers via the expanded form.
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1)
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + .
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1)
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – .
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving Equations
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving EquationsThe most important application for factoring is to solve polynomial equations.
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving EquationsThe most important application for factoring is to solve polynomial equations. These are equations of the form
polynomial = polynomial
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving EquationsThe most important application for factoring is to solve polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving EquationsThe most important application for factoring is to solve polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.Fact: If A*B = 0,then either A = 0 or B = 0
Example C. Determine the outcome is + or – for x2 – 2x – 3 if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)Hence for x = -3/2:(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving EquationsThe most important application for factoring is to solve polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.Fact: If A*B = 0,then either A = 0 or B = 0For example, if 3x = 0, then x must be equal to 0.
Example D. a. If 3(x – 2) = 0,
Applications of Factoring
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0,
Applications of Factoring
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0,
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0,
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor(x – 3)(x + 1) = 0
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor(x – 3)(x + 1) = 0
There are two linear x–factors. We may extract one answer from each.
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor(x – 3)(x + 1) = 0Hence x – 3 = 0 or x + 1 = 0
There are two linear x–factors. We may extract one answer from each.
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor(x – 3)(x + 1) = 0Hence x – 3 = 0 or x + 1 = 0 x = 3
There are two linear x–factors. We may extract one answer from each.
Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. 2. factor the polynomial,3. get the answers.
b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor(x – 3)(x + 1) = 0Hence x – 3 = 0 or x + 1 = 0 x = 3 or x = -1
There are two linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Applications of Factoring
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x
Applications of Factoring
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3
Applications of Factoring
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0
Applications of Factoring
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0
Applications of Factoring
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0
Applications of Factoring
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0
Applications of Factoring
or x – 1 = 0
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3
Applications of Factoring
or x – 1 = 0
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = -3
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = -3 x = -3/2
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = -3 2x = 3 x = -3/2
There are three linear x–factors. We may extract one answer from each.
b. 2x(x + 1) = 4x + 3(1 – x) Expand 2x2 + 2x = 4x + 3 – 3x 2x2 + 2x = x + 3 Set one side 0 2x2 + 2x – x – 3 = 0 2x2 + x – 3 = 0 Factor (2x + 3)(x – 1) = 0 Get the answers 2x + 3 = 0 2x = -3 x = -3/2
Applications of Factoring
or x – 1 = 0 x = 1
c. 8x(x2 – 1) = 10x Expand 8x3 – 8x = 10x Set one side 0 8x3 – 8x – 10x = 0 8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0 x = 0 or 2x + 3 = 0 or 2x – 3 = 0 2x = -3 2x = 3 x = -3/2 x = 3/2
There are three linear x–factors. We may extract one answer from each.
Exercise A. Use the factored form to evaluate the following expressions with the given input values.
Applications of Factoring
1. x2 – 3x – 4, x = –2, 3, 5 2. x2 – 2x – 15, x = –1, 4, 7
3. x2 – 2x – 1, x = ½ ,–2, –½ 4. x3 – 2x2, x = –2, 2, 4
5. x3 – 4x2 – 5x, x = –4, 2, 6 6. 2x3 – 3x2 + x, x = –3, 3, 5
B. Determine if the output is positive or negative using the factored form.7. x2 – 3x – 4, x = –2½, –2/3, 2½, 5¼
8. –x2 + 2x + 8, x = –2½, –2/3, 2½, 5¼
9. x3 – 2x2 – 8x, x = –4½, –3/4, ¼, 6¼,
11. 4x2 – x3, x = –1.22, 0.87, 3.22, 4.01
12. 18x – 2x3, x = –4.90, –2.19, 1.53, 3.01
10. 2x3 – 3x2 – 2x, x = –2½, –3/4, ¼, 3¼,
C. Solve the following equations. Check the answers.
Applications of Factoring
18. x2 – 3x = 10 20. x(x – 2) = 24 21. 2x2 = 3(x + 1) – 1
28. x3 – 2x2 = 0
22. x2 = 4
25. 2x(x – 3) + 4 = 2x – 4
29. x3 – 2x2 – 8x = 0
31. 4x2 = x3
30. 2x2(x – 3) = –4x
26. x(x – 3) + x + 6 = 2x2 + 3x
13. x2 – 3x – 4 = 0 14. x2 – 2x – 15 = 0 15. x2 + 7x + 12 = 0
16. –x2 – 2x + 8 = 0 17. 9 – x2 = 0 18. 2x2 – x – 1 = 0
27. x(x + 4) + 9 = 2(2 – x)
23. 8x2 = 2 24. 27x2 – 12 = 0
32. 4x = x3 33. 4x2 = x4
34. 7x2 = –4x3 – 3x 35. 5 = (x + 2)(2x + 1)
36. (x – 1)2 = (x + 1)2 – 4 37. (x + 1)2 = x2 + (x – 1)2
38. (x + 2)2 – (x + 1)2= x2 39. (x + 3)2 – (x + 2)2 = (x + 1)2
