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CHAPTER 6 Factoring
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10: x12 1 x6 1 x6 1 x3 1 x3 1 x6 1 When the rst term is not x2 , see if you can factor out the coe cient of x2 If you can, then you are left with a quadratic whose rst term is x2 For example each term in 2x2 16x 18 is divisible by 2: 2x2 16x 18 2 x2 8x 9 2 x 9 x 1
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1: 4x2 28x 48 2: 3x2 9x 54 3: 9x2 9x 18 4: 15x2 60 5: 6x2 24x 24
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1: 4x2 28x 48 4 x2 7x 12 4 x 4 x 3 2: 3x2 9x 54 3 x2 3x 18 3 x 6 x 3 3: 9x2 9x 18 9 x2 x 2 9 x 2 x 1 4: 15x2 60 15 x2 4 15 x 2 x 2 5: 6x2 24x 24 6 x2 4x 4 6 x 2 x 2 6 x 2 2 The coe cient of the x2 term will not always factor away In order to factor quadratics such as 4x2 8x 3 you will need to try all combinations of factors of 4 and of 3: 4x x and 2x 2x The blanks will be lled in with the factors of 3 You will need to check all of the possibilities: 4x 1 x 3 , 4x 3 x 1 , and 2x 1 2x 3 :
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4x2 4x 15
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The possibilities to check are (a) (c) (e) (g) (i) 4x 15 x 1 4x 1 x 15 4x 5 x 3 4x 3 x 5 2x 15 2x 1 (f)
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CHAPTER 6 Factoring
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(b) 4x 15 x 1 (d) 4x 1 x 15 4x 5 x 3
(h) 4x 3 x 5 (j) (l) 2x 15 2x 1 2x 5 2x 3
(k) 2x 5 2x 3
We have chosen these combinations to force the rst and last terms of the quadratic to be 4x2 and 15, respectively, we only need to check the combination that will give a middle term of 4x (if there is one) (a) (c) (e) (g) (i) 4x 15x 11x 60x x 59x 12x 5x 7x 20x 3x 17x 2x 30x 28x (b) 4x 15x 11x (d) 60x x 59x (f) 12x 5x 7x
(h) 20x 3x 17x (j) (l) 2x 30x 28x 6x 10x 4x
(k) 6x 10x 4x
Combination (l) is the correct factorization: 4x2 4x 15 2x 5 2x 3 : You can see that when the constant term and x2 s coe cient have many factors, this list of factorizations to check can grow rather long Fortunately there is a way around this problem as we shall see in a later chapter
Practice
1: 6x2 25x 9 2: 18x2 21x 5 3: 8x2 35x 12
CHAPTER 6 Factoring
4: 25x2 25x 14 5: 4x2 9 6: 4x2 20x 25 7: 12x2 32x 35
Solutions
1: 6x2 25x 9 2x 9 3x 1 2: 18x2 21x 5 3x 1 6x 5 3: 8x2 35x 12 8x 3 x 4 4: 25x2 25x 14 5x 2 5x 7 5: 4x2 9 2x 3 2x 3 6: 4x2 20x 25 2x 5 2x 5 2x 5 2 7: 12x2 32x 35 6x 5 2x 7
Quadratic Type Expressions
An expression with three terms where the power of the rst term is twice that of the second and the third term is a constant is called a quadratic type expression They factor in the same way as quadratic polynomials The power on x in the factorization will be the power on x in the middle term To see the e ect of changing the exponents, let us look at x2 2x 3 x 3 x 1 : x4 2x2 3 x2 3 x2 1 x6 2x3 3 x3 3 x3 1 x10 2x5 3 x5 3 x5 1
x 4 2x 2 3 x 2 3 x 2 1 x2=3 2x1=3 3 x1=3 3 x1=3 1 x1 2x1=2 3 x1=2 3 x1=2 1
CHAPTER 6 Factoring
Examples
4x6 20x3 21 2x3 3 2x3 7 x2=3 5x1=3 6 x1=3 2 x1=3 3 x4 x2 2 x2 2 x2 1 x2 2 x 1 x 1 p x 2 x 8 x1 2x1=2 8 x1=2 4 x1=2 2 p p x 4 x 2 p p x 2 4 x 15 x1=2 2x1=4 15 x1=4 5 x1=4 3 p p 4 x 5 4 x 3
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