Intermediate Algebra: A Graphing Approach

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Intermediate Algebra A Graphing Approach

• 5.8 Solving Equations by Factoring
  Problem Solving

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Section 5.8

• Polynomial equations
• Equations that set 2 polynomials equal to each
  other.
• Standard form has a 0 on one side of the
  equation.
• Quadratic equations
• Polynomial equations of degree 2.
• Zero factor theorem
• If a and b are real numbers and ab 0, then a
  0 or b 0.
• This property is true for three or more factors,
  as well.

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• Steps for solving a polynomial equation by
  factoring
• Write the equation in standard form.
• Factor the polynomial completely.
• Set each factor containing a variable equal to 0.
• Solve the resulting equations.
• Check each solution in the original equation.

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• Solve x2 5x 24.
• First write the polynomial equation in standard
  form.
• x2 5x 24 0
• Now we factor the polynomial using techniques
  from the previous sections.
• x2 5x 24 (x 8)(x 3) 0
• We set each factor equal to 0.
• x 8 0 or x 3 0, which will simplify
  to
• x 8 or x -3

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• Check both possible answers in the original
  equation.
• 82 5(8) 64 40 24 true
• (-3)2 5(-3) 9 (-15) 24 true
• So our solutions for x are 8 or 3.

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• Solve 4x(8x 9) 5
• First write the polynomial equation in standard
  form.
• 32x2 36x 5
• 32x2 36x 5 0
• Now we factor the polynomial using techniques
  from the previous sections.
• 32x2 36x 5 (8x 1)(4x 5) 0
• We set each factor equal to 0.
• 8x 1 0 or 4x 5 0

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• Check both possible answers in the original
  equation.

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• General strategy for problem solving
• Understand the problem
• Read and reread the problem
• Choose a variable to represent the unknown
• Construct a drawing, whenever possible
• Propose a solution and check
• Translate the problem into an equation
• Solve the equation
• Interpret the result
• Check proposed solution in problem
• State your conclusion

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The product of two consecutive positive integers
is 132. Find the two integers.
Read and reread the problem. If we let x
one of the unknown positive integers, then
x 1 the next consecutive positive integer.
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x(x 1) 132
x2 x 132 (distributive property)
x2 x 132 0 (write quadratic in
standard form)
(x 12)(x 11) 0 (factor quadratic
polynomial)
x 12 0 or x 11 0 (set factors
equal to 0)
x -12 or x 11 (solve each factor for
x)
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Check Remember that x is suppose to represent a
positive integer. So, although x -12 satisfies
our equation, it cannot be a solution for the
problem we were presented. If we let x 11, then
x 1 12. The product of the two numbers is 11
12 132, our desired result. State The two
positive integers are 11 and 12.
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• Pythagorean Theorem
• In a right triangle, the sum of the squares of
  the lengths of the two legs is equal to the
  square of the length of the hypotenuse.
• (leg a)2 (leg b)2 (hypotenuse)2

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Find the length of the shorter leg of a right
triangle if the longer leg is 10 miles more than
the shorter leg and the hypotenuse is 10 miles
less than twice the shorter leg.
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Read and reread the problem. If we let x
the length of the shorter leg, then x 10
the length of the longer leg and 2x 10
the length of the hypotenuse.
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By the Pythagorean Theorem, (leg a)2 (leg b)2
(hypotenuse)2 x2 (x 10)2 (2x
10)2
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x2 (x 10)2 (2x 10)2
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Check Remember that x is suppose to represent
the length of the shorter side. So, although x
0 satisfies our equation, it cannot be a solution
for the problem we were presented. If we let x
30, then x 10 40 and 2x 10 50. Since 302
402 900 1600 2500 502, the Pythagorean
Theorem checks out. State The length of the
shorter leg is 30 miles. (Remember that is all
we were asked for in this problem.)
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• Recall that previously, we found the x-intercept
  of linear equations by letting y 0 and solving
  for x.
• The same method works for x-intercepts in
  quadratic equations.
• Note When the quadratic equation is written in
  standard form, the graph is a parabola opening up
  (when a gt 0) or down (when a lt 0), where a is the
  coefficient of the x2 term.
• The x-intercepts will be where the parabola
  crosses the x-axis.

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• Find the x-intercepts of the graph of y 4x2
  11x 6.
• The equation is already written in standard form,
  so we let y 0, then factor the quadratic in x.
• 0 4x2 11x 6 (4x 3)(x 2)
• We set each factor equal to 0 and solve for x.
• 4x 3 0 or x 2 0
• 4x -3 or x -2
• x -¾ or x -2
• So the x-intercepts are the points (-¾, 0) and
  (-2, 0).