5 7applications of factoring
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5 7applications of factoring 5 7applications of factoring Presentation Transcript

  • Applications of Factoring
  • Applications of Factoring There are many applications of the factored forms of polynomials.
  • Applications of Factoring There are many applications of the factored forms of polynomials. Following are some of them:
  • Applications of Factoring There are many applications of the factored forms of polynomials. Following are some of them: 1. to evaluate polynomials,
  • Applications of Factoring There 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 Factoring There 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 Factoring There 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 Factoring There 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 Factoring There 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. Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first.
  • Applications of Factoring There 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. Example A. Evaluate x2 – 2x – 3 if x = 7 a. without factoring. b. by factoring it first. We get 72 – 2(7) – 3
  • Applications of Factoring There 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. 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
  • Applications of Factoring There 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. 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
  • Applications of Factoring There 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. 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)
  • Applications of Factoring There 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. 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)
  • Applications of Factoring There 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. 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
  • Applications of Factoring There 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. 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
  • Applications of Factoring 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 Your turn: Double check these answers via the expanded form.
  • 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 Your turn: Double check these answers via the expanded form. Determine the signs of the outputs
  • 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 Your turn: Double check these answers via the expanded form. Determine the signs of the outputs Often we only want to know the sign of the output, i.e. whether the output is positive or negative.
  • 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 Your turn: Double check these answers via the expanded form. Determine the signs of the outputs Often 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.
  • Applications of Factoring 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. Solving Equations
  • 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. Solving Equations The most important application for factoring is to solve polynomial equations.
  • 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. Solving Equations The most important application for factoring is to solve polynomial equations. These are equations of the form polynomial = polynomial
  • 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. Solving Equations The 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.
  • 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. Solving Equations The 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
  • 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. Solving Equations The 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 For example, if 3x = 0, then x must be equal to 0.
  • Applications of Factoring 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. b. If (x + 1)(x – 2) = 0,
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0,
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 To solve polynomial equation,
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side. x = –1 or x = 2
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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,
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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.
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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. Example E. Solve for x a. x2 – 2x – 3 = 0
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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. Example E. Solve for x a. x2 – 2x – 3 = 0 Factor (x – 3)(x + 1) = 0
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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. 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.
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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. Example E. Solve for x a. x2 – 2x – 3 = 0 Factor (x – 3)(x + 1) = 0 Hence x – 3 = 0 or x + 1 = 0 There are two linear x–factors. We may extract one answer from each.
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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. Example E. Solve for x a. x2 – 2x – 3 = 0 Factor (x – 3)(x + 1) = 0 Hence x – 3 = 0 or x + 1 = 0 x = 3 There are two linear x–factors. We may extract one answer from each.
  • Applications of Factoring Example D. a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2. b. If (x + 1)(x – 2) = 0, then either (x + 1) = 0 or (x – 2) = 0, x = –1 or x = 2 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. Example E. Solve for x a. x2 – 2x – 3 = 0 Factor (x – 3)(x + 1) = 0 Hence 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.
  • Applications of Factoring 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 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 There are three linear x–factors. We may extract one answer from each. 2x = -3 2x = 3 x = -3/2 x = 3/2
  • Applications of Factoring Exercise A. Use the factored form to evaluate the following expressions with the given input values. 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, 10. 2x3 – 3x2 – 2x, x = –2½, –3/4, ¼, 3¼, 11. 4x2 – x3, x = –1.22, 0.87, 3.22, 4.01 12. 18x – 2x3, x = –4.90, –2.19, 1.53, 3.01
  • Applications of Factoring C. Solve the following equations. Check the answers. 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 18. x2 – 3x = 10 20. x(x – 2) = 24 21. 2x2 = 3(x + 1) – 1 23. 8x2 = 2 24. 27x2 – 12 = 0 26. x(x – 3) + x + 6 = 2x2 + 3x 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 27. x(x + 4) + 9 = 2(2 – x) 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