Transcript No Slide Title

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Slide 7- 1
5.5
Equations
Containing
Trinomials of
the Type
ax2 + bx + c
■
Factoring Trinomials of the Type
ax2 + bx + c
■
Equations and Functions
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METHOD 1: Factoring with FOIL
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To Factor ax2 + bx + c using FOIL
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Example Factor: 3x2  14x  5
Solution
1. First, check for a common factor. There is
none other than 1 or 1.
2. Find the First terms whose product is 3x2.
The only possibilities are 3x and x:
(3x + )(x + )
3. Find the Last terms whose product is 5.
Possibilities are (5)(1), (5)(1)
Important!: Since the First terms are not identical,
we must also consider the above factors in reverse
order: (1)(5), and (1)(5).
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Slide 5- 5
continued
Factor: 3x2  14x  5
4. Knowing that the First and Last products will check, inspect
the Outer and Inner products resulting from steps (2) and (3)
Look for the combination in which the sum of the products is
the middle term.
(3x  5)(x + 1) = 3x2 + 3x  5x  5
Wrong middle term
= 3x2  2x  5
(3x  1)(x + 5) = 3x2 + 15x  x  5
Wrong middle term
= 3x2 + 14x  5
(3x + 5)(x  1) = 3x2  3x + 5x  5
Wrong middle term
= 3x2 + 2x  5
(3x + 1)(x  5) = 3x2  15x + x  5
= 3x2  14x  5
Correct middle term!
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Example Factor: 18x3  141x2  24x
Solution
1. First, we factor out the largest common factor, 3x:
3x(6x2  47x  8)
2. Factor 6x2  47x  8. Since 6x2 can be factored as
3x  2x or 6x x, we have two possibilities
(3x + )(2x + ) or (6x + )(x + )
3. There are several pairs of factors of 8. List each
way:
8, 1
1, 8
2, 4
4, 2
8, 1
1, 8
2, 4
4, 2
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Slide 5- 7
continued Factor: 3x(6x2  47x  8)
Trial
Product
(6x + 4)(x  2) = 6x2  12x + 4x  8
= 6x2  8x  8
Wrong middle term
(6x  4)(x + 2) = 6x2 + 12x  4x  8
= 6x2 + 8x + 8
Wrong middle term
(6x + 1)(x  8) = 6x2  48x + x  8
Correct middle term
= 6x2  47x  8
We do not need to consider (3x + )(2x + ).
The
complete factorization is 3x(6x + 1)(x  8).
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Example Factor: 14x + 5  3x2
Solution
It is an important problem-solving strategy to find a
way to make problems look like problems we already
know how to solve. Rewrite the equation in
descending order.
14x + 5  3x2 = 3x2 + 14x + 5
Factor out the 1:
3x2 + 14x + 5 = 1(3x2 14x  5)
= 1(3x + 1)(x  5)
The factorization of 14x + 5  3x2 is 1(3x + 1)(x  5).
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Slide 5- 9
Tips for Factoring ax2 + bx + c with FOIL
1. If the largest common factor has been factored out of
the original trinomial, then no binomial factor can have
a common factor (other than 1 or –1).
2. If a and c are both positive, then the signs in the factors
will be the same as the sign of b.
3. When a possible factoring produces the opposite of the
desired middle term, reverse the signs of the constants
in the factors.
4. Be systematic about your trials. Keep track of those
possibilities that you have tried and those that you have
not.
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Slide 5- 10
METHOD 2: The ac-Method
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To Factor ax2 + bx + c, Using the ac-Method
1. Make sure that any common factors have been
factored out.
2. Multiply the leading coefficient a and the
constant c.
3. Find a pair of factors p and q, such that pq = ac
and p + q = b.
4. Rewrite the middle term of the trinomial, bx, as
px + qx.
5. Factor by grouping.
6. Include any common factor from step (1) and
check by multiplying.
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Example Factor 4x2  5x  6
Solution
1. First, we note that there is no common factor (other
than 1 or 1).
2. Multiply the leading coefficient, 4 and the
constant, 6:
(4)(6) = 24.
3. Try to factor 24 so that the sum of the factors is 5:
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continued Factor 4x2  5x  6
3. Pairs of Factors
of 24
1, 24
1, 24
2, 12
2, 12
3, 8
3, 8
4, 6
4, 6
Sums of
Factors
23
23
10
10
5
5
2
2
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We would normally
stop listing pairs of
factors once we have
found the one we are
after.
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continued Factor 4x2  5x  6
4. Split 5x using the results of step (3):
5x = 8x + 3x.
5. Factor by grouping:
4x2  5x  6 = 4x2  8x + 3x  6
= 4x(x  2) + 3(x  2)
= (x  2)(4x + 3)
We check the solution by multiplying or using a table.
Check: (x  2)(4x + 3) = 4x2 + 3x  8x  6
= 4x2  5x  6
The factorization of 4x2  5x  6 is (x  2)(4x + 3).
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Example Factor 8x3 + 10x2  12x
Solution
1. We factor out the greatest common factor, 2x:
8x3 + 10x2  12x = 2x(4x2 + 5x  6)
2. To factor 4x2 + 5x  6 by grouping, we multiply
the leading coefficient, 4 and the constant term
(6):
4(6) = 24.
3. We next look for pairs of
Pairs of
Sums of
factors of 24 whose
Factors of
Factors
24
sum is 5.
3, 8
3, 8
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5
5
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continued Factor 8x3 + 10x2  12x
4. We then rewrite the middle term: 4x2 + 5x  6 using
5x = 3x + 8x
5. Factor by grouping:
4x2 + 5x  6 = 4x2  3x + 8x  6
= x(4x  3) + 2(4x  3)
= (x + 2)(4x  3)
6. The factorization of the original trinomial
8x3 + 10x2  12x = 2x(x + 2)(4x  3).
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Slide 5- 17
Equations and Functions
We now use our new factoring skills to solve a
polynomial equation. We factor a polynomial
expression and use the principle of zero products to
solve the equation.
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Slide 5- 18
Example
Solve: 2x5 – 7x4 + 3x3 = 0.
Solution--Algebraic
We note that the polynomial has degree 5, so there will
be at most 5 solutions of the equation.
2x5 – 7x4 + 3x3 = 0
x3(2x – 1)(x – 3) = 0
Factoring
x3 = 0
or 2x – 1 = 0
or x – 3 = 0
x=0
or
x=½
or
x=3
The solutions are 0, ½ and 3.
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continued--Graphical
We find the x-intercepts of the function 2x5 – 7x4 + 3x3
using the ZERO option of the CALC menu.
(0, 0)
(0.5, 0)
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(3, 0)
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Example
Find the domain of F if F(x) =
x 1
x2  2x  8
Solution
The domain of F is the set of all values for which the
function is a real number. Since division by 0 is
undefined, we exclude any x-value for which the
denominator is 0.
x2 + 2x – 8 = 0
(x – 2)(x + 4) = 0
x–2=0
or x + 4 = 0
x=2
or
x = –4 These are the values to exclude.
The domain of F is {x|x is a real number and x 2
and x  4}.
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