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6.3
Trinomials of the Type ax2 + bx + c
■ Factoring Trinomials of the Type ax2 + bx + c
■ Equations and Functions
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METHOD 2: The ac-Method
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To Factor ax2 + bx + c, Using the
ac-Method
1. Factor out the largest common factor, if one exists.
2. Multiply the leading coefficient a and the constant c.
3. Find a pair of factors ac whose sum is b.
4. Rewrite the middle term as a sum or a difference using the factors found in step (3).
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
Sums of Factors
1, 24
23
1, 24 23
2, 12
10
2, 12 10
3, 8 5
3, 8 5
4, 6 2
4, 6 2
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
factors of 24 whose
sum is 5.
Pairs of Factors of 24
Sums of Factors
3, 8 5
3, 8 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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9. 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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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) = Factoring
x3 = or 2x – 1 = or x – 3 = 0
x = or x = ½ or x = 3
The solutions are 0, ½ and 3.
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11. 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)
(3, 0)
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Example
Find the domain of F if F(x) =
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 = or x + 4 = 0
x = 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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