Factoring Quadratic Equations

10 Key excerpts on "Factoring Quadratic Equations"

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  Algebra

  Form and Function

  • William G. McCallum, Eric Connally, Deborah Hughes-Hallett(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  In the previous section we looked at two ways of expressing a quadratic: in factored form and in vertex form. Each way leads to a method for solving the corresponding quadratic equation. Solving Quadratic Equations by Factoring In Example 1 on page 124 we considered the quadratic function which gives the revenue of selling T-shirts at price  dollars, () = (1000 − 20). Since this function is already given as a product of linear factors, we can find its zeros by finding where the factors are zero. In general, we have: The Zero-Factor Principle If  ⋅  = 0 then either  = 0 or  = 0 (or both). According to the zero-factor principle, the only possibilities for  ⏟ ⏟ ⏟  (1000 − 20) ⏟⏞⏞⏞⏞⏞ ⏟⏞⏞⏞⏞⏞ ⏟  = 0 are  = 0 or 1000 − 20 = 0. This gives the solutions  = 0 and  = 50. If we can factor a quadratic equation written in standard form easily, then we can use the zero- factor principle to find the solutions. Example 1 Solve the quadratic equations using the zero-factor principle and factoring. (a)  2 − 4 + 3 = 0 (b) 3 2 + 3 = 6 3.4 QUADRATIC EQUATIONS 117 Solution (a) We have  2 − 4 + 3 = ( − 1)( − 3) = 0, so we want to solve the equation ( − 1) ⏟ ⏟ ⏟  ( − 3) ⏟ ⏟ ⏟  = 0. If  is a solution, then either  = 0, which implies  − 1 = 0  = 1, or  = 0, which implies  − 3 = 0  = 3. (b) In order to use the zero-factor principle, we must first write the quadratic equation in the standard form by subtracting 6 from both sides. This gives 3 2 + 3 − 6 = 0  2 +  − 2 = 0 divide both sides by 3 ( + 2)( − 1) = 0. This implies  = −2 and  = 1 are the solutions. In general, the zero-factor principle tells us that for a quadratic equation in the form ( − )( − ) = 0, where , ,  are constants,  ≠ 0, the only solutions are  =  and  = . Solving Quadratic Equations with Perfect Squares Factoring is a useful technique for solving quadratic equations, but sometimes factoring may be difficult (or impossible).

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  Technical Mathematics with Calculus

  • Michael A. Calter, Paul A. Calter, Paul Wraight, Sarah White(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  Solving quadratics has been enhanced by the evo- lution of today’s scientific calculators. 14–1 Solving Quadratics by Factoring Terminology A polynomial equation of second degree is called a quadratic equation. It is common practice to refer to it simply as a quadratic. Recall that in a polynomial, all powers of x are positive integers. 14 Quadratic Equations 303 Section 14–1 ◆ Solving Quadratics by Factoring ◆◆◆ Example 1: The following equations are quadratic equations: (a) 4x 2 - 5x + 2 = 0 (c) 9x 2 - 5x = 0 (b) x 2 = 58 (d) 2x 2 - 7 = 0 Equation (a) is called a complete quadratic; (b) and (d), which have no x terms, are called pure quadratics; and (c), which has no constant term, is called an incomplete quadratic. A quadratic is in general form when it is written in the following form, where a, b, and c are constants: General Form of a Quadratic ax 2 + bx + c = 0 99 ◆◆◆ Example 2: Write the following quadratic equation in general form, then identify the values of a, b, and c, as in Eq. 99, above. (Manipulate the equation.) x x 7 4 5 3 2 - = Solution: Since fractions are generally unpopular when manipulating equations, let’s move the 3 from the fraction x 5 3 2 by multiplying both sides by 3. This allows us to get rid of the 3 in the denominator. Step 1: x x (3) (7 4 ) 5 3 (3) 2 × - = × Let’s expand the left side of the equation. Step 2: 3 . (7 - 4x) = 5x 2 Your equation should look like this so far. Step 3: 21 - 12x = 5x 2 Now move all terms to the left side of the equation in descending order, starting with the x 2 term. Be sure place a zero on the right side of the equation. Step 4: -5x 2 - 12x + 21 = 0 One of the math rules for quadratics is to make sure that the first term is positive, so we can multiply the whole equation by -1. Step 5: (-1) × (-5x 2 - 12x + 21 = 0) We now have the original equation in general form. 5x 2 + 12x - 21 = 0 The equation is now in general form, with a = 5, b = 12, and c = -21.

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  Foundations of Mathematics

  Algebra, Geometry, Trigonometry and Calculus

  • Philip Brown(Author)
  • 2016(Publication Date)
  • Mercury Learning and Information

    (Publisher)

  Because any quadratic polynomial has two roots, it can be expressed as a product of two linear factors multiplied by a constant. As a demonstration in table 3.5, we give the factorized form of each of the quadratic polynomials from table 3.3. 68 • Foundations of Mathematics TABLE 3.5. Quadratic polynomials in factorized form Quadratic polynomials Factorized form (i) 2 4 6 2 x x − − 2 1 3 ( )( ) x x + − (ii) − − − x x 2 4 4 − + ( ) x 2 2 (iii) 6 15 2 x x − − +       -       x x 6 3 2 5 3 (iv) x x 2 11 − − - +         - -         x x 1 45 2 1 45 2 (v) − − x 2 2 − − + ( 2 )( 2 ) x x i i (vi) + + x x 5 2 2 - +       - -       x x 1 3i 2 1 3i 2 EXAMPLE 3.5.3. The factorized form for 6 15 2 x x − − (row (iii)) can also be expressed as 6 15 2 3 2 3 5 3 2 3 3 5 2 x x x x x x − − = +       −       = + − ( )( ).  Finding the factors of a polynomial is called factorizing or factoring (we will use the former). It is possible to factorize quadratic polynomials with integer coefficients by inspection if they factorize into a pair of linear factors that also have integer coefficients (e.g., the polynomials in rows (i) and (iii) of the table above). This is a very useful skill, and it is also fun! The trick is to find the correct pairs of factors for the leading coefficient and the constant term. For example, for the polynomial 6 15 2 x x − − (row (iii) in the table), the correct pair of factors of the leading coefficient is 3 and 2 and the correct pair of factors of the constant term is 3 and 5. Finding the correct pairs of factors can be done by trial and error, but it does help to be methodical. We demonstrate this with four sets of examples. EXAMPLE 3.5.4. Factorize, by inspection, each of the following quadratic polynomials.

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  Beginning Algebra

  Connecting Concepts through Applications

  • Mark Clark, Cynthia Anfinson(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Sometimes we can solve these equations by setting one side equal to 0 and then factoring. The zero-product property applies to any product that is equal to 0. Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. S E C T I O N 6 . 4 S o l v i n g Q u a d r a t i c E q u a t i o n s b y F a c t o r i n g 573 6.4 Vocabulary Exercises 1. A quadratic equation is a polynomial equation with degree _______________ . 2. An equation of the form ax 2 1 bx 1 c 5 0 is called the _______________ of a quadratic equation. 3. When two factors are multiplied together to get 0, the _______________ states that one or both of the two must equal 0. 6.4 Exercises For Exercises 1 through 14, determine whether each equation is linear, quadratic, or neither. Do not solve the equations. 1. 2 x 1 5 5 3 2. 7 t 1 2 5 11 3. 5 x 3 1 7 x 2 2 8 5 0 4. x 3 1 6 x 5 9 5. x 2 1 2 x 2 4 5 0 6. 3 x 2 2 6 x 1 7 5 0 7. 4 x 2 1 3 x 2 2 5 0 8. 7 b 2 2 5 b 1 6 5 0 9. 6 x 5 12 10. 3 1 x 1 1 2 5 2 12 11. 5 1 x 1 2 2 5 3 x 2 8 12. 2 4 1 x 1 1 2 5 x 13. 5 1 2 y 5 4 y 2 14. 7 5 8 h 2 1 3 h For Exercises 15 through 28, solve each equation using the zero-product property. Check the answers. 15. 5 1 x 1 3 2 5 0 16. 8 1 x 2 6 2 5 0 17. 2 4 1 2 n 2 5 2 5 0 18. 2 2 1 3 p 1 1 2 5 0 19. a 1 a 2 8 2 5 0 20. m 1 m 2 4 2 5 0 21. y 1 2 y 1 7 2 5 0 22. x 1 3 x 1 4 2 5 0 23. 3 t 1 2 t 1 5 2 5 0 24. 2 h 1 5 h 1 12 2 5 0 25. 1 x 1 5 21 x 2 6 2 5 0 26. 5 1 x 1 2 2 1 x 2 3 2 5 0 27. 4 p 1 2 p 1 3 21 p 2 9 2 5 0 28. 2 k 1 5 k 2 4 21 2 k 1 3 2 5 0 For Exercises 29 through 46, solve each equation by factoring.

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  Years 9 - 10 Maths For Students

  • (Author)
  • 2015(Publication Date)
  • For Dummies

    (Publisher)

  Fractions reduce more easily, equations solve more easily, and answers are observed more easily when you can factor. 180 Part II: Algebra is Part of Everything Factoring out numbers Factoring is the opposite of distributing; it’s ‘undistributing’ (refer to Chapter 7 for more on distribution). When performing distribution, you multiply a series of terms by a common multiplier. Now, by factoring, you seek to find what a series of terms has in common and then take it away, dividing the common factor or multiplier out from each term. Think of each term as a numerator of a fraction, and you’re finding the same denominator for each. By factoring out, the common factor is put outside parentheses or brackets and all the results of the divisions are left inside. An expression can be written as the product of the largest number that divides all the terms evenly times the results of the divisions: ab + ac + ad = a(b + c + d ). Writing factoring as division In the trinomial 16a − 8b + 40c 2 , 2 is a common factor. But 4 is also a common factor, and 8 is a common factor. Here are the divisions of the terms by 2, 4 and 8: Reviewing the terms and rules You’ll understand factoring better if you have a firm handle on what the terms used to talk about factoring mean: 6 Term: A group of number(s) and/or varia- ble(s) connected to one another by multi- plication or division and separated from other terms by addition or subtraction. 6 Factor: Any of the values involved in a multiplication problem that, when multi- plied together, produce a result. 6 Coefficient: A number that multiplies a variable and tells how many of the variable. 6 Constant: A number or variable that never changes in value. 6 Relatively prime: Terms that have no factors in common. If the only factor that numbers share in common is 1, they’re considered relatively prime. Here is an illustration for all the terms I just gave: In the expression 5xy + 4z − 6, you see three terms.

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  eBook - ePub

  Elementary Algebra

  • Toby Wagner(Author)
  • 2021(Publication Date)
  • Chemeketa Press

    (Publisher)

  It’s time to learn how to use factoring to help us solve quadratic equations. To do this, we need to make use of the zero-product property, and in order to use the zero-product property, one side of the equation must be 0. With that in mind, we have the following three-step process.

  Solving Quadratic Equations by Factoring

  1.   To use the zero-product property, one side of the equation must be 0. If necessary, use algebra to put all nonzero terms on one side of the equal sign and 0 on the other side, thus rewriting the equation in standard form: a x 2 + bx + c = 0.

  2.   Factor the quadratic expression: ( )( ) = 0

  3.   Use the zero-product property to solve the equation.

  We will use this process to work through the following examples.

  Example 4

  Solve x 2 – 7x + 12 = 0

  Solution

  Using mental math, we write the solutions: x = 3, and x = 4.

  We check our work by substituting each value back into the original equation.

  In the examples that follow, our focus will be on solving equations. Therefore, we won’t show the work that goes in to checking the solutions. However, in practice, it is always a good idea to check the solutions you find when solving an equation.

  Example 5

  Solve the following equations.

  1.     x2 = 25

  2.     x2 = 2x

  Solutions

  1.     x2 = 25

  2.     x 2 = 2x

  In part 2 of the previous example, the first factor, x, was a monomial. We need to be careful, however, to not write a solution of 0 every time we see a monomial factor. If a monomial factor does not include a variable, then we cannot write 0 as a solution.

  Example 6

  Solve each equation.

  1.     8 x2 = 16x + 24

  2.     7 x3 + 63 x2 = 154x

  Solutions

  1.

      8 x 2 = 16x + 24

  It’s typically easier to solve quadratic equations in which the quadratic polynomial has a positive leading coefficient. The term 8 x 2 has a positive coefficient, so we’ll leave it alone.

  2.

      7 x 3 + 63 x 2 = 154x

  This is not a quadratic equation. The highest power of the variable is 3, so this is a third-degree or cubic

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  Beginning and Intermediate Algebra

  A Guided Approach

  • Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Equations such as 9 x 2 2 6 x 5 0 and 3 x 2 1 4 x 2 7 5 0 that contain second-degree polynomials are called quadratic equations . QUADRATIC EQUATIONS A quadratic equation in one variable is an equation of the form ax 2 1 bx 1 c 5 0 (This is called quadratic form .) where a , b , and c are real numbers, and a 2 0 . Solve a quadratic equation in one variable using the zero-factor property. Many quadratic equations can be solved by factoring. For example, to solve the quadratic equation x 2 1 5 x 2 6 5 0 which is already in quadratic form, we begin by factoring the trinomial and writing the equation as 1 x 1 6 21 x 2 1 2 5 0 1 (1) Copyright 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 5.7 Solving Equations by Factoring 347 ZERO-FACTOR PROPERTY Suppose a and b represent two real numbers. If ab 5 0 , then a 5 0 or b 5 0 . Comment In mathematics, when we use the word “or” it is understood to mean one or the other or both. Thus, if a ? b 5 0 , both a and b could be equal to 0. By applying the zero-factor property to Equation 1, we have x 1 6 5 0 or x 2 1 5 0 We can solve each of these linear equations to get x 5 2 6 or x 5 1 To check, we substitute 2 6 for x , and then 1 for x in the original equation and simplify. For x 5 2 6 For x 5 1 x 2 1 5 x 2 6 5 0 x 2 1 5 x 2 6 5 0 1 2 6 2 2 1 5 1 2 6 2 2 6 0 0 1 1 2 2 1 5 1 1 2 2 6 0 0 36 2 30 2 6 0 0 1 1 5 2 6 0 0 6 2 6 0 0 6 2 6 0 0 0 5 0 0 5 0 Both solutions check. The quadratic equations 9 x 2 2 6 x 5 0 and 4 x 2 2 25 5 0 are each missing a term.

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  Elementary Algebra

  • Alan Tussy, R. Gustafson(Authors)
  • 2012(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  A polynomial is factored completely when no factor can be factored further. Factor: Is there a common factor? No. There is no common factor (other than 1). How many terms does it have? Since the polynomial has four terms, try factoring by grouping. Factor from and 4 from . Factor out . Is it factored completely? No. We can factor as a sum of two cubes. is prime. is prime. Does it check? Use multiplication to check. a 2 2 a 4 ( a 2)( a 2 2 a 4) ( a 2 4) a 2 4 ( a 3 8) ( a 2 4) a 5 8 a 2 4 a 3 32 a 3 8 a 3 8 ( a 3 8) ( a 2 4) 4 a 3 32 a 5 8 a 2 a 2 a 5 8 a 2 4 a 3 32 a 2 ( a 3 8) 4 ( a 3 8) a 5 8 a 2 4 a 3 32 REVIEW EXERCISES Factor. 55. 56. 57. 58. 59. 60. 12 w 4 36 w 3 27 w 2 400 x 400 m 2 x m 2 3 j 3 24 j 4 16 5 s 2 t 5 s 2 u 2 5 tv 5 u 2 v 14 y 3 6 y 4 40 y 2 61. 62. 63. 64. 18 c 3 d 2 12 c 3 d 24 c 2 d x 2 z 64 y 2 z 16 xyz 121 p 2 36 q 2 2 t 3 10 SECTION 6.7 Solving Quadratic Equations by Factoring DEFINITIONS AND CONCEPTS EXAMPLES A quadratic equation is an equation that can be written in the standard form , where , , and are real numbers and . a 0 c b a ax 2 bx c 0 Examples of quadratic equations are: 5 x 2 25 x 0, 4 a 2 9 0, and y 2 13 y 6 The Zero-Factor Property If the product of two (or more) numbers is 0, then at least one of the numbers is 0. If then, or . x 3 0 x 2 0 ( x 2 )( x 3 ) 0 To use the factoring method to solve a quadratic equation: 1. Write the equation in standard form: or 2. Factor completely. 3. Use the zero-factor property to set each factor equal to 0. 4. Solve each resulting equation. 5. Check each result in the original equation. 0 ax 2 bx c ax 2 bx c 0 Solve: The solutions are 0 and . The solution set is {0, }. Check each result in the original equation. 5 5 x 0 x 5 5 x 0 or x 5 0 5 x ( x 5 ) 0 5 x 2 25 x 0 Solve: The solutions are and . The solution set is { , } . 3 2 3 2 3 2 3 2 a 3 2 a 3 2 2 a 3 2 a 3 2 a 3 0 or 2 a 3 0 ( 2 a 3 )( 2 a 3 ) 0 4 a 2 9 0 Copyright 201 Cengage Learning.

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  Introductory Algebra

  Concepts with Applications

  • Charles P. McKeague(Author)
  • 2013(Publication Date)
  • XYZ Textbooks

    (Publisher)

  427 5.7 Solving Quadratic Equations by Factoring Step 5: Substitute each solution into 2x 2 − 5x = 12 to check. Check: − 3 __ 2 Check: 4 2  − 3 __ 2  2 − 5  − 3 __ 2  ≟ 12 2(4) 2 − 5(4) ≟ 12 2  9 __ 4  + 5  3 __ 2  = 12 2(16) − 20 = 12 9 __ 2 + 15 __ 2 = 12 32 − 20 = 12 24 __ 2 = 12 12 = 12 12 = 12 EXAMPLE 2 Solve the equation 16a 2 − 25 = 0. Solution The equation is already in standard form. 16a 2 − 25 = 0 (4a − 5)(4a + 5) = 0 Factor the left side. 4a − 5 = 0 or 4a + 5 = 0 Set each factor equal to 0. 4a = 5 4a = −5 Solve the resulting equations. a = 5 __ 4 a = − 5 __ 4 EXAMPLE 3 Solve the equation 4x 2 = 8x. Solution We begin by adding −8x to each side of the equation to put it in standard form. Then we factor the left side of the equation by factoring out the greatest common factor. 4x 2 = 8x 4x 2 − 8x = 0 Add −8x to each side. 4x(x − 2) = 0 Factor the left side. 4x = 0 or x − 2 = 0 Set each factor equal to 0. x = 0 x = 2 Solve the resulting equations. The solutions are 0 and 2. EXAMPLE 4 Solve the equation x(2x + 3) = 44. Solution We must multiply out the left side first and then put the equation in standard form. x(2x + 3) = 44 2x 2 + 3x = 44 Multiply out the left side. 2x 2 + 3x − 44 = 0 Add −44 to each side. (2x + 11)(x − 4) = 0 Factor the left side. 2x + 11 = 0 or x − 4 = 0 Set each factor equal to 0. 2x = −11 x = 4 Solve the resulting equations. x = − 11 __ 2 The two solutions are − 11 __ 2 and 4. 2. Solve the equation 49a 2 − 16 = 0. Answers 2. − 4 _ 7 , 4 _ 7 3. 0, − 2 _ 5 4. 5, 8 3. Solve the equation 5x 2 = −2x. 4. Solve the equation x(13 − x) = 40. Chapter 5 Factoring Polynomials 428 EXAMPLE 5 Solve the equation 5 2 = x 2 + (x + 1) 2 . Solution Before we can put this equation in standard form we must square the binomial. Remember, to square a binomial, we use the formula (a + b ) 2 = a 2 + 2ab + b 2 . 5 2 = x 2 + (x + 1) 2 25 = x 2 + x 2 + 2x + 1 Expand 5 2 and (x + 1) 2 .

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  Intermediate Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  To generalize the preceding example, here are the steps used in solving an equation by factoring. Solve: 100 x 2 = 300 x . SOLUTION We begin by writing the equation in standard form and factoring. 100 x 2 = 300 x 100 x 2 − 300 x = 0 Standard form 100 x ( x − 3) = 0 Factor Using the zero-factor property to set each factor to 0, we have 100 x = 0 or x − 3 = 0 x = 0 x = 3 The two solutions are 0 and 3. Solve: ( x − 2)( x + 1) = 4. SOLUTION We begin by multiplying the two factors on the left side. (Notice that it would be incorrect to set each of the factors on the left side equal to 4. The fact that the product is 4 does not imply that either of the factors must be 4.) ( x − 2)( x + 1) = 4 x 2 − x − 2 = 4 Multiply the left side x 2 − x − 6 = 0 Standard form ( x − 3)( x + 2) = 0 Factor x − 3 = 0 or x + 2 = 0 Zero-factor property x = 3 x = − 2 Note We are placing a question mark over the equal sign because we don’t know yet if the expression on the left will be equal to the expression on the right. Step 1: Write the equation in standard form. Step 2: Factor the left side. Step 3: Use the zero-factor property to set each factor equal to 0. Step 4: Solve the resulting linear equations. HOW TO Solve an Equation by Factoring EXAMPLE 2 EXAMPLE 3 5.8 Solving Equations by Factoring 395 Solve: ( x + 2)(3 x − 1) = ( x + 2)( x + 6). SOLUTION We begin by multiplying the factors on each side. ( x + 2)(3 x − 1) = ( x + 2)( x + 6) 3 x 2 + 5 x − 2 = x 2 + 8 x + 12 FOIL both sides 2 x 2 − 3 x − 14 = 0 Standard form (2 x − 7)( x + 2) = 0 Factor 2 x − 7 = 0 or x + 2 = 0 Zero-factor Property 2 x = 7 x = − 2 x = 7 _ 2 We have two solutions: 7 _ 2 and − 2. Solving Polynomial Equations by Factoring We can use the zero-factor property to solve polynomial equations of higher degree. The process is similar to the method we have used to solve quadratic equations. Solve: x 3 + 2 x 2 − 9 x − 18 = 0. SOLUTION We start with factoring by grouping.

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