8-2 Factoring Using the Distributive Property Recall using the distributive property for multiplying a monomial by a polynomial:

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x(a + b) = ax + bx Doing this produces a common factor in each term. In the above case that common factor is x. We can use this fact to help us factor polynomials that have a common factor in all of their terms. For example, the terms in the polynomial:

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12a2 +16a have a common factor, this common factor is

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4a (it is the GCF). We can factor this polynomial by “undoing” the distributive property and factor out the GCF:

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12a2 +16a = 4a(3a + 4) We can check our answer by simply using the distributive property to multiply. Example: Factor

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18cd2 +12c 2d + 9cd

1. The GCF of all the terms is

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3cd , so divide all of the terms by

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3cd 2.

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3cd(6d + 4c + 3) 3. Check by multiplying:

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3cd(6d + 4c + 3) =18cd2 +12c 2d + 9cd , this is correct. Using Grouping to Factor Sometimes all of the terms in a polynomial do not have a GCF, but pairs of terms do. When this is the case we can use grouping to factor. Example: Factor

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4ab + 8b + 3a + 6 Notice that all of the terms do not have a GCF, but the first two and the last two terms do. We can group them in the following way:

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(4ab + 8b) + (3a + 6) We can now factor each binomial separately:

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4b(a + 2) + 3(a + 2) Now notice that each has a factor of

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(a + 2), factor again:

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(a + 2)(4b + 3) Done.

Example: Factor

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6x 2 −15x − 8x + 20 We can group this polynomial like this:

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(6x 2 −15x) + (−8x + 20) Factor each binomial:

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3x(2x − 5) + −4(2x − 5) Notice that I factored a -4 out of the second binomial instead of a 4, why? So the resulting binomial would match the first one. Notice that each now has a factor of

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(2x − 5), factor again:

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(2x − 5)(3x − 4) Check the answer by multiplying using the FOIL method:

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2x(3x) − 2x(4) − 5(3x) + 5(4) = 6x 2 − 8x −15x + 20, this is correct! F O I L Example: Factor

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rs+ 5s − r − 5 We can group this polynomial like this:

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(rs − r) + (5s − 5) {I rearranged the terms} Now factor each binomial:

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r(s −1) + 5(s −1) Notice that each now has a factor of

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(s −1), factor again:

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(s −1)(r + 5) Check the answer by multiplying using the FOIL method:

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rs+ 5s −1r − 5 = rs+ 5s − r − 5 , this is correct! F O I L Example: Factor

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35x − 5xy + 3y − 21 We can group this polynomial like this:

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(35x − 5xy) + (3y − 21) Now factor each binomial:

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5x(7 − y) + 3(y − 7) Notice that the resulting binomials are not the same, but they are almost the same…we can make them the same by factoring a -1 out of the first binomial:

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5x(−1)(−7 + y) + 3(y − 7) = −5x(y − 7) + 3(y − 7) Now the binomials match, factor again:

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(y − 7)(−5x + 3) This is correct.

Example: Factor

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c − 2cd + 8d − 4

1. Group the terms:

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(c − 2cd) + (8d − 4) 2. Factor each binomial:

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c(1− 2d) + 4(2d −1) 3. Factor a -1 out of the first binomial:

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c(−1)(−1+ 2d) + 4(2d −1) 4. Simplify

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−c(2d −1) + 4(2d −1) 5. Factor out (2d-1)

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(2d −1)(−c + 4) Example: Factor

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3p − 2p2 −18p + 27

1. Group the terms:

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(3p − 2p2) + (−18p + 27) 2. Factor each binomial:

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p(3 − 2p) + 9(−2p + 3) 3. Rearrange the terms:

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p(3 − 2p) + 9(3 − 2p) 4. Factor out (3-2p)

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(3 − 2p)(p + 9) Solving Equations by Factoring: We can use factoring to help us solve equations. To do this we will use the Zero Product Property. The Zero Product Property states that if two things multiply to get zero, one or both of them MUST be equal to zero. If

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ab = 0 then either

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a = 0, b = 0, or a& b = 0 The solutions to an equation are called roots. For example, if we have the equation

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x(x −1) = 0 we can say one of two things: 1.

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x = 0 or 2.

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x −1 = 0 So we can say that this equation has two solutions, or roots. Either

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x = 0 or x =1

Example: Solve the equation

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(d − 5)(3d + 4) = 0 Because of the Zero Product Property (ZPP) we can say one of two things: 1.

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d − 5 = 0 2.

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3d + 4 = 0 Letʼs solve both of these scenarios to find the roots

1.

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d − 5 = 0d = 5

2.

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3d + 4 = 03d = −4

d = −43

Therefore,

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d = 5 or − 43

Example: Solve the equation

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x 2 = 7x In order to use the ZPP, weʼll have to set one side of the equation to zero:

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x 2 − 7x = 0 Now factor the binomial:

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x(x − 7) = 0 Now we can use the ZPP to say one of two things: 1.

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x = 0 2.

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x − 7 = 0 Letʼs solve both of the scenarios to find the roots 1.

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x = 0 2.

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x − 7 = 0x = 7

Therefore,

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x = 0 or x = 7

Example: Solve the equation

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x 2 = −10x In order to use the ZPP, weʼll have to set one side of the equation to zero:

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x 2 +10x = 0 Now factor the binomial:

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x(x +10) = 0 Now we can use the ZPP to say that 1.

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x = 0 2.

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x +10 = 0 Now find the roots 1.

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x = 0 2.

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x +10 = 0x = −10

Therefore,

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x = 0 or −10