Solving Quadratic Equations by Factoring 5 2 Solving Quadratic Equations by mikesanye

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									 5.2: Solving Quadratic Equations
 by Factoring
 (p. 256)

Tucker Method of Factoring
     Factoring Simple Trinomials

   Factoring simple polynomials in
    the form of x2 + bx + c.          x2 + 5x + 6   Factors of:
                                                         +6
   An example would be:
    ◦ x2 + 5x + 6                      x     x       3   ●     2
                                                    -3   ●    -2
    ◦ First factor the last                          6   ●     1
      number , +6 (include                          -6   ●    -1

      the sign).
    ◦ Next, write the first
      term (x2) without the
      exponent on the top
      of the bar.
     Factoring Simple Trinomials

   Factoring x2 + 5x + 6              x2 + 5x + 6   Factors of:
    ◦ First factor the last number ,                    +6
      +6 (include the sign).
                                        x     x
    ◦ Find the factors of that                         3 + 2
                                                     -3 + -2
                                                               = +5
                                                               = -5
      number (+6), that the            +3    +2       6 + 1
                                                     -6 + -1
                                                               = +7
                                                               = -7
      sum (+) results in the
      middle number. (+5).
    ◦ Place those factors (+3
      and +2) in the box
      under the bar.
     Factoring Simple Trinomials

   Factoring x2 + 5x + 6          x2 + 5x + 6
    ◦ Now take the left side
                                                   Factors of:
                                                      +6
      (x + 3) and place that
      binomial in a parenthesis.    x     x          3 + 2 = +5
                                                     6 + 1 = +7
    ◦ Then take the right side     +3    +2
      (x + 2) and place that
      binomial in a
      parenthesis.                         (x + 3) (x + 2) = x2 + 5x + 6

    ◦ You have now factored
      the polynomial.
     Factoring Simple Trinomials

   Now, Factor: x2 - 5x - 6
                                x2 - 5x - 6
    ◦ First factor the last                   Factors of:
      number , -6 (include                       -6
      the sign).                  x     x     -3 +    2    = -1
    ◦ Find the factors of                      3 +    -2   = +1
      that number (-6),         -6    +1       6 +    -1   = +5
      that the difference (-)                 -6 +     1   = -5

      results in the middle
      number. (-5).
    ◦ Place those factors (-
      6 and +1) in the box
      under the bar.
     Factoring Simple Trinomials

   Factoring x2 - 5x - 6      x2 - 5x - 6
    ◦ Now take the left
                                                  Factors of:
                                                       -6
      side (x - 6) and place
      that binomial in a         x       x        -3   +     2   = -1

      parenthesis.             -6       +1
                                                   3
                                                   6
                                                       +
                                                       +
                                                            -2
                                                            -1
                                                                 = +1
                                                                 = +5
    ◦ Then take the right                         -6   +     1   = -5

      side (x + 1) and place
      that binomial in a
      parenthesis.                   (x - 6) (x + 1) = x2 - 5x - 6
    ◦ You have now
      factored the
      polynomial.
     Factoring Simple Trinomials

   Factor: x2 - 7x + 12
                               x2 - 7x +12
    ◦ First factor the last                  Factors of:
      number , +12                             +12
      (include the sign).       x     x       2 + 6 = +8
    ◦ Find the factors of                    -2 + -6 = -8
      that number (+12),       -3    -4       3 + 4 = +7
      that the sum (+)                       -3 + -4 = -7
      results in the middle                  12 + 1 = +13
      number. (-7).                          -6 + 1 = -5

    ◦ Place those factors (-
      3 and -4) in the box
      under the bar.
     Factoring Simple Trinomials

   Factor: x2 - 7x + 12       x2 - 7x + 12
    ◦ Now take the left
                                              Factors of:
                                                 +12
      side       (x - 3) and
      place that binomial       x      x       2 + 6 = +8
      in a parenthesis.                       -2 + -6 = -8
                                                3 + 4 = +7
    ◦ Then take the right       -3    -4       -3 + -4 = -7
      side (x - 4) and place                   12 + 1 = +13
                                               -6 + 1 = -5
      that binomial in a
      parenthesis.              (x - 3) (x - 4) = x2 - 7x + 12
    ◦ You have now
      factored the
      polynomial.
    Factoring Trinomials
     ax2 + bx + c

 Factoring trinomials in the
                                      5x2 - x -18
  form of: ax2 + bx + c.                            Factors of:
                                                      5 ● -18 = -90
 An example would be:
                                        5x    5x
   ◦ 5x2 - x –18                                              1 ●
                                                               2 ●
                                                                     -90
                                                                     -45
   ◦ First, place the first term on                            3 ●   -20
                                                               5 ●   -18
     both sides of the bars.                                   6 ●   -15
                                                               9 ●
     (without the exponent)                                   10 ●
                                                                     -10
                                                                     -9
   ◦ Second, multiply the first                               Etc……
     (5) and the last number
     (-18).
   ◦ Find the factors of -90.
     Factoring Trinomials
     ax2 + bx + c

   Factoring trinomials in the
                                   5x2 - x -18
    form of: ax2 + bx + c.                       Factors of:
                                                    -90
 Factor: 5x2    - x –18
                                     5x    5x     1 + -90 = -89
    ◦ After finding the factors,                  2 + -45 = - 44
      choose the set whose            9    -10    3 + -20 = -17
                                                  5 + -18 = - 13
      difference equals the                       6 + -15 = -9
      middle number, -1.                          9 + -10 = -1
      (Remember that –x is the                   10 + -9 = +1
      same as –1x).                              Etc……

    ◦ Place that set with signs
      under the bar.
     Factoring Trinomials
     ax2 + bx + c

   Factoring trinomials in the
                                  5x2 - x -18
    form of: ax2 + bx + c.                         Factors of:
                                                      -90
   To finish factoring you
                                     5x   5x
    must reduce each side of                        1 + -90 = -89
                                                    2 + -45 = - 44
    the bar if possible.             9     -10      3 + -20 = -17
                                                    5 + -18 = - 13
   The 5x over –10 will                  1x        6 + -15 = -9

                                          2
                                                    9 + -10 = -1
    reduce to 1x (or x) over -                     10 + -9 = +1
    2.                                             Etc……
                                  (5x + 9) (x - 2) = 5x2 - x –18
   Now, move each side into
    the factored answer.
     Factoring Trinomials
      ax2 + bx + c

   Another example would
    be:                        9x2 + 18x -16   Factors of:
                                                9 ● -16 = -144
    ◦ 9x2 +18x – 16
    ◦ First, place the first       9x    9x              12 ● -12
                                                         4 ● -38
      term on both sides of                              3 ● -48
      the bars. (without the                             6 ● -24
                                                        -6 ● 24
      exponent)                                         -4 ● 36
                                                        -2 ● 72
    ◦ Second, multiply the                              Etc……
      first (9) and the last
      number (-16).
    ◦ Find the factors.
     Factoring Trinomials
     ax2 + bx + c

   Factoring trinomials in
                                 9x2 + 18x -16
    the form of: ax2 + bx + c.                   Factors of:
                                                    -144
   Factor: 9x2 + 18x –16            9x    9x    12 + -12 = 0
    ◦ After finding the                           4 + -38 = -34

      factors, choose the            -6   +24     3 + -48 = -45
                                                  6 + -24 = -18
      set whose difference                       -6 + 24 = +18
                                                 -4 + 36 = +32
      equals the middle                          -2 + 72 = +70
                                                  Etc……
      number, +18.
    ◦ Place that set with
      signs under the bar.
     Factoring Trinomials
     ax2 + bx + c

   Factor: 9x2 + 18x –16
                               9x2 + 18x -16
   To finish factoring you                      Factors of:
                                                    -144
    must reduce each side of
    the bar if possible.           9x    9x       12 + -12 = 0
                                                   4 + -38 = -34
   The 9x over –6 will            -6    +24       3 + -48 = -45
                                                   6 + -24 = -18
    reduce to 3x over -2.           3x     3x     -6 + 24 = +18
                                                  -4 + 38 = +34
   The 9x over +24 will            -2     +8     -2 + 72 = +70
                                                   Etc……
    reduce to 3x over +8.
   Now, move each side into    (3x -2) (3x +8) = 9x2 +18x –16
    the factored answer.
     Factoring Trinomials
      ax2 + bx + c

   The last example contains a
    ‘Greatest Common Factor’:       2t2 + 3t - 5
     ◦ 6t3 + 9t2 – 15t
     ◦ First factor out the
       Greatest Common Factor
       (GCF) which is 3t.
     ◦ Go ahead and place the
       GCF in your answer so you
       do not forget it.
     ◦ Now, place the remaining       3t(   )(     ) = 6t3 + 9t2 – 15t
       trinomial in your box with
       the bars.
     Factoring Trinomials
      ax2 + bx + c

   The last example contains a
    ‘Greatest Common Factor:            2t2 + 3t - 5                Factors of:
     ◦ 6t3 + 9t2 – 15t = 3t(2t2 +3t –                                  -10 = 2 ● -5
       5)
                                            2t       2t               1 + -10 = -9
     ◦ Next, place the first term                                     2 + -5 = -3
       on both sides of the bars.               5        -2           5 + -2 = +3
       (without the exponent)                                         10 + -1 = +9
     ◦ Then, multiply the first
       number (2) and the last
       number (-5).
     ◦ Find the factors that the
                                          3t(       )(        ) = 6t3 + 9t2 – 15t
       difference (-) will give you
       the middle term (+3). Place
       those factors under the bar.
     Factoring Trinomials
      ax2 + bx + c

   To finish factoring you     2t2 + 3t - 5
    must reduce each side of                          Factors of:
                                                         -10 = 2 ● -5
    the bar if possible.
                                      2t 2t             1 + -10 = -9
   The left side of the bar                            2 + -5 = -3
                                      5    -2
    does not reduce, but the                            5 + -2 = +3
                                                        10 + -1 = +9
    right side does reduce to              1t
                                           -1
    1t (or t) over –1.
   Now, move each side into
                                3t(2t +5)(t - 1) = 6t3 + 9t2 – 15t
    the factored answer.
Factoring Trinomials
Tucker Method
Factor these trinomials completely:
  1. 6y2 + 13y + 6
  2. 15t2 + 19t – 10
  3. 18x2 – 24x + 6
  4. 12x2 – 28x – 24
  5. 15x3 + 19x2 – 10x
Hand the answers into your instructor in class.
    Zero Product Property


 Let A and B be real numbers or algebraic
  expressions. If AB=0, then A=0 or B=0.
 This means that If the product of 2 factors
  is zero, then at least one of the 2 factors
  had to be zero itself!
 Ex: If xy=0, then either x=0 or y=0.
 Example: Solve.
 2t2-17t+45=3t-5

2t2-17t+45=3t-5    Set eqn. =0
2t2-20t+50=0       factor out GCF of 2
2(t2-10t+25)=0     divide by 2
t2-10t+25=0        factor left side
(t-5)2=0           set factors =0
t-5=0              solve for t
+5 +5
t=5                check your solution!
Example: Solve.
x2+3x-18=0

x2+3x-18=0          Factor the left side
(x+6)(x-3)=0        set each factor =0
x+6=0 OR x-3=0      solve each eqn.
 -6 -6      +3 +3
  x=-6 OR x=3       check your solutions!
  Example: Solve.
  3x-6=x2-10

3x-6=x2-10           Set = 0
0=x2-3x-4            Factor the right side
0=(x-4)(x+1)         Set each factor =0
x-4=0 OR x+1=0       Solve each eqn.
 +4 +4       -1 -1
x=4 OR x=-1          Check your solutions!
Finding the Zeros of an Equation
   The Zeros of an equation are the x-
    intercepts !

 First, change y to a zero.
 Now, solve for x.
 The solutions will be the zeros of the
  equation.
 Example: Find the Zeros of
 y=x2-x-6
y=x2-x-6                Change y to 0
0=x2-x-6                Factor the right side
0=(x-3)(x+2)            Set factors =0
x-3=0 OR x+2=0          Solve each equation
 +3 +3      -2 -2
 x=3 OR x=-2            Check your solutions!

If you were to graph the eqn., the graph would
   cross the x-axis at (-2,0) and (3,0).
Assignment

								
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