Copyright 2013, 2009, 2006 Pearson Education, Inc. 1 Section 6.6
Solving Quadratic Equations by Factoring Copyright 2013, 2009, 2006
Pearson Education, Inc. 1 Slide 2 2 Slide 3 3 Solving Polynomial
Equations We have spent much time on learning how to factor
polynomials. Now we will look at one important use of factoring. In
this section, we will use factoring to solve equations of degree 2.
Up to this point, we have only looked at solving equations of
degree one. Slide 4 Copyright 2013, 2009, 2006 Pearson Education,
Inc. 4 Solving Quadratic Equations Definition of a Quadratic
Equation A quadratic equation in x is an equation that can be
written in the standard form where a, b, and c are real numbers,
with. A quadratic equation in x is also called a second-degree
polynomial equation in x. The Zero-Product Rule If the product of
two algebraic expressions is zero, then at least one of the factors
is equal to zero. If AB = 0, then A = 0 or B = 0. Slide 5 Copyright
2013, 2009, 2006 Pearson Education, Inc. 5 Solve the equation (x +
5) (2x 1) = 0. By the zero-product principle, the only way this
product can be zero is if at least one of the factors is zero.
Thus, x + 5 = 0 or2x 1 = 0 x = 5 2x = 1 x = The two solutions are 5
and . Solving Quadratic EquationsEXAMPLE Slide 6 Copyright 2013,
2009, 2006 Pearson Education, Inc. 6 Solve the equation (x + 5) (2x
1) = 0. By the zero-product principle, the only way this product
can be zero is if at least one of the factors is zero. Thus, x + 5
= 0 or2x 1 = 0 x = 5 2x = 1 x = The two solutions are 5 and .
Solving Quadratic EquationsEXAMPLE Slide 7 Copyright 2013, 2009,
2006 Pearson Education, Inc. 7 Objective #1: Example Slide 8
Copyright 2013, 2009, 2006 Pearson Education, Inc. 8 Objective #1:
Example Slide 9 Copyright 2013, 2009, 2006 Pearson Education, Inc.
9 Slide 10 10 Solving Polynomial Equations Solving a Quadratic
Equation by Factoring 1) If necessary, rewrite the equation in the
standard form, moving all terms to one side, thereby obtaining zero
on the other side. 2) Factor. 3) Apply the zero-product principle,
setting each factor equal to zero. 4) Solve the equations formed in
step 3. 5) Check the solutions in the original equation. Slide 11
Copyright 2013, 2009, 2006 Pearson Education, Inc. 11 Solve x 2 7x
= 8 Step 1: Rewrite the equation in standard form ax 2 + bx + c =
0. x 2 7x = 8 x 2 7x 8 = 0 Step 2: Factor: x 2 7x 8 = 0 (x 8) (x
+1) = 0 Solving Quadratic Equations by FactoringEXAMPLE Slide 12
Copyright 2013, 2009, 2006 Pearson Education, Inc. 12 Solve x 2 7x
= 8 Step 1: Rewrite the equation in standard form ax 2 + bx + c =
0. x 2 7x = 8 x 2 7x 8 = 0 Step 2: Factor: x 2 7x 8 = 0 (x 8) (x
+1) = 0 Solving Quadratic Equations by FactoringEXAMPLE Slide 13
Copyright 2013, 2009, 2006 Pearson Education, Inc. 13 Solve x 2 7x
= 8 by factoring Steps 3 and 4: Set each factor equal to zero and
solve. x 8 = 0 or x + 1 = 0 x = 8 or x = 1 Step 5: Check the
solutions in the original equation. 8 2 7(8) = 8(1) 2 7(1) = 8 64
56 = 8 1 + 7 = 8 8 = 8 8 = 8 Solving Quadratic Equations by
FactoringCONTINUED Slide 14 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 14 Solve x 2 7x = 8 by factoring Steps 3 and 4: Set
each factor equal to zero and solve. x 8 = 0 or x + 1 = 0 x = 8 or
x = 1 Step 5: Check the solutions in the original equation. 8 2
7(8) = 8(1) 2 7(1) = 8 64 56 = 8 1 + 7 = 8 8 = 8 8 = 8 Solving
Quadratic Equations by FactoringCONTINUED Slide 15 Copyright 2013,
2009, 2006 Pearson Education, Inc. 15 Solving Polynomial
EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and
obtain zero on the other side. Subtract 45 from both sides and
write the equation in standard form. Subtract 45 from both sides
Simplify 2) Factor. Factor Slide 16 Copyright 2013, 2009, 2006
Pearson Education, Inc. 16 Solving Polynomial Equations 3) & 4)
Set each factor equal to zero and solve the resulting equations. or
CONTINUED 5) Check the solutions in the original equation. ? ? ? ?
Slide 17 Copyright 2013, 2009, 2006 Pearson Education, Inc. 17
Solving Polynomial EquationsCONTINUED The solutions are 9 and 5.
The solution set is {9, 5}. ?? The graph of is shown here. true
Slide 25 Copyright 2013, 2009, 2006 Pearson Education, Inc. 25
Objective #2: Example Slide 26 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 26 Objective #2: Example Slide 27 Copyright 2013,
2009, 2006 Pearson Education, Inc. 27 Objective #2: Example Slide
28 Copyright 2013, 2009, 2006 Pearson Education, Inc. 28 Objective
#2: Example Slide 29 Copyright 2013, 2009, 2006 Pearson Education,
Inc. 29 Objective #2: ExampleCONTINUED Slide 30 Copyright 2013,
2009, 2006 Pearson Education, Inc. 30 Objective #2:
ExampleCONTINUED Slide 31 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 31 Objective #2: Example Slide 32 Copyright 2013,
2009, 2006 Pearson Education, Inc. 32 Objective #2: Example Slide
33 Copyright 2013, 2009, 2006 Pearson Education, Inc. 33 Objective
#2: ExampleCONTINUED Slide 34 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 34 Objective #2: ExampleCONTINUED Slide 35
Copyright 2013, 2009, 2006 Pearson Education, Inc. 35 Objective #2:
Example Slide 36 Copyright 2013, 2009, 2006 Pearson Education, Inc.
36 Objective #2: Example Slide 37 Copyright 2013, 2009, 2006
Pearson Education, Inc. 37 Objective #2: ExampleCONTINUED Slide 38
Copyright 2013, 2009, 2006 Pearson Education, Inc. 38 Slide 39
Copyright 2013, 2009, 2006 Pearson Education, Inc. 39 Polynomial
Equations in ApplicationEXAMPLE A gymnast dismounts the uneven
parallel bars at a height of 8 feet with an initial upward velocity
of 8 feet per second. The function describes the height of the
gymnasts feet above the ground, s (t), in feet, t seconds after
dismounting. The graph of the function is shown below. Slide 40
Copyright 2013, 2009, 2006 Pearson Education, Inc. 40 Polynomial
Equations in ApplicationSOLUTION When will the gymnast be 8 feet
above the ground? Identify the solution(s) as one or more points on
the graph. We note that the graph of the equation passes through
the line y = 8 twice. Once when x = 0 and once when x = 0.5. This
can be verified by determining when y = s (t) = 8. That is,
CONTINUED Original equation Replace s (t) with 8 Subtract 8 from
both sides Factor Slide 41 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 41 Polynomial Equations in Application Now we set
each factor equal to zero. CONTINUED We have just verified the
information we deduced from the graph. That is, the gymnast will
indeed be 8 feet off the ground at t = 0 seconds and at t = 0.5
seconds. These solutions are identified by the dots on the graph on
the next slide. or Slide 42 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 42 Polynomial Equations in ApplicationCONTINUED
Slide 43 Copyright 2013, 2009, 2006 Pearson Education, Inc. 43
Objective #3: Example Slide 44 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 44 Objective #3: Example Slide 45 Copyright 2013,
2009, 2006 Pearson Education, Inc. 45 Objective #3:
ExampleCONTINUED Slide 46 Copyright 2013, 2009, 2006 Pearson
Education, Inc. 46 Objective #3: ExampleCONTINUED
