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Sometimes when looking for the greatest common factor, you may find that it is 1. 18a 5 b 2 + 21a 7 + 14b 6 GCF of 18, 21 and 14 is 1 Not all terms have the variable a so it is not common to all terms Not all terms have the variable b so it is not common to all terms GCF = 1 Factors: 1 (18a 5 b 2 + 21a 7 + 14b 6 ) There is nothing that is edifying when you remove a factor of 1. The leftover terms are identical to the original polynomial. There is no point in removing a common factor of 1.

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Sometimes when you find that the GCF is 1, there are other ways to find factors of the polynomial. However, whenever you have to factor a polynomial, you must at least determine if there is a GCF and remove it if there is. When the GCF is 1, one of the other methods that you may use to factor a polynomial is grouping as with this next example. ab + bc + ad + cd The GCF of the numerical coefficients is 1, not all terms have the variable a, variable b, variable c nor variable d. That means there is no point in removing a GCF from each term. In this case, we can separate the terms of the polynomial into 2 groups. ab + bc GROUP 1 + ad + cd GROUP 2

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ab + bc GROUP 1 + ad + cd GROUP 2 Instead of searching for a common factor for all 4 terms we can search for the GCF of each group. ab + bc GROUP 1 The GCF for group 1 is b. b(a + c) GROUP 1 ad + cd GROUP 2 The GCF for group 1 is d. d(a + c) GROUP 2 ab + bc+ ad + cd = b(a + c)+ d(a + c) b(a + c)+ d(a + c) Notice that the leftover factors for each group are the same ‘a + c’. Also each of the factored groups is now a term of the expression and we would say that it has only 2 terms (instead of 4) in its transformed state. Those 2 terms have a common binomial factor. ‘a + c’ is a common factor in both terms. b(a + c) d(a + c) = +

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= (a + c)(b + d) We can now factor this transformed polynomial by dividing each term by the GCF of a + c. GCF = a + c b (a + c)+ d (a + c) Review of the steps: ab + bc + ad + cd ab + bc+ ad + cd= b (a + c)+ d (a + c)= = (a + c)(b + d) d(a + c) TERM 2 + b(a + c) TERM 1

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bx + 2x GROUP 1 + ab + 2a GROUP 2 bx + 2x + ab + 2aExample 2: The GCF for group 1 is x. x(b + 2) GROUP 1 a(b + 2) GROUP 2 bx + 2x + ab + 2a = x(b + 2) + a(b + 2) The 2 terms (instead of 4) of the transformed state of the polynomial have a common binomial factor. That factor (GCF) is b + 2. = x(b + 2) + a(b + 2) = (b + 2) (x + a) The GCF for group 2 is a.

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x 2 + ab GROUP 1 + ax + bx GROUP 2 x 2 + ab + ax + bxExample 3: The GCF for group 1 is 1. 1(x 2 + ab) GROUP 1 x(a + b) GROUP 2 The 2 terms of the polynomial have a different binomial factor. Grouping will not work with the polynomial expressed in this form. However, if we change the order of the terms and try again, it can be made to work. The GCF for group 2 is x. We can switch the second and third terms and retry. + bx+ ax+ abx2x2 + bx+ ax + ab x2x2 + bx + ab x2x2 + ax + bxx2x2 + ax + ab The GCF for group 1 is x. x(x + a) GROUP 1 b(a + x) GROUP 2 The GCF for group 2 is b. (x + a) = (a + x) We can write these binomials either way. + ab+ bxx2x2 + ax x 2 + ax GROUP 1 + ab + bx GROUP 2

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x 2 + ab + ax + bxExample 3: x 2 + ax GROUP 1 + ab + bx GROUP 2 = x(x + a) + b(a + x) The 2 new terms have a common binomial factor. That factor (GCF) is x + a. = x(x + a) + b(x + a) = (x + a) (x + b) x(x + a) GROUP 1 + b(a + x) GROUP 2 = x(x + a) + b(x + a) ab - 2ac + 3ad + 4b – 8c + 12dExample 4: The previous examples have all been polynomials with 4 terms but this example has 6 terms. There are two ways to organize the groups to factor this. We can make 2 groups of three terms or 3 groups of two terms. We will factor this polynomial both ways.

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ab - 2ac + 3ad + 4b – 8c + 12d Making 2 groups of three terms: To make these 2 groups we must identify some commonality amongst each group of three terms. The first 3 terms contain the variable a and the last 3 terms are all factors of 4 plus each of the last 3 have only 1 variable. Based on this, we can form the 2 groups without changing the order of the terms. ab - 2ac + 3ad GROUP 1 + 4b – 8c + 12d GROUP 2 The GCF for group 1 is a.The GCF for group 2 is 4. a(b - 2c + 3d) GROUP 1 4(b – 2c + 3d) GROUP 2 The leftover terms for each group are identical and can be used as the GCF for the final step of factoring.

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= ab - 2ac + 3ad GROUP 1 + 4b – 8c + 12d GROUP 2 = a(b - 2c + 3d)+ 4(b – 2c + 3d) ab - 2ac + 3ad + 4b – 8c + 12dThe 2 new terms have a common trinomial factor. That factor (GCF) is b – 2c + 3d. = a(b - 2c + 3d)+ 4(b – 2c + 3d) When each of the 2 new terms a divided by the common trinomial factor, the leftover terms are a and +4. = (a + 4) (b - 2c + 3d) There are 2 factors of the of the original polynomial and when they are multiplied, we will get the polynomial that we started with. This polynomial can be factored by making 3 groups of two terms instead.

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ab - 2ac + 3ad + 4b – 8c + 12d Making 3 groups of two terms: To make these 3 groups we must identify some commonality to form each group of two terms. First, there are two terms that contain the variable b, then two terms that contain the variable c and finally two terms contain the variable d. Based on this, we can from the 3 groups by changing the order of the terms in the polynomial that we are trying to factor. - 2ac – 8c GROUP 2 The GCF for group 1 is b. The GCF for group 2 is -2c. b(a + 4) -2c(a + 4) ab - 2ac + 3ad + 4b – 8c + 12d = ab - 2ac + 3ad + 4b – 8c + 12d + 3ad + 12d GROUP 3 ab + 4b GROUP 1 = The GCF for group 3 is 3d. 3d(a + 4) = b(a + 4)- 2c(a + 4)+ 3d(a + 4) Notice, when factoring group 2, that the leading term is negative. (- 2ac), therefore the GCF is also negative. (-2c)

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The leftover terms for all 3 groups are identical and can be used as the GCF for the final step of factoring. This involves dividing each of the 3 factored groups by that GCF. - 2ac – 8c GROUP 2 ab - 2ac + 3ad + 4b – 8c + 12d = ab - 2ac + 3ad + 4b – 8c + 12d + 3ad + 12d GROUP 3 ab + 4b GROUP 1 = = b(a + 4)- 2c(a + 4)+ 3d(a + 4) = b(a + 4)- 2c(a + 4)+ 3d(a + 4) = (b – 2c + 3d) (a + 4)

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ac - ad + 4bd – 4bcExample 5: The GCF for all 4 terms is 1. We must try to factor by grouping. We can group the terms containing variable a and the terms containing 4 or variable b. The GCF for group 1 is a. The GCF for group 2 is 4b. ac - ad GROUP 1 + 4bd – 4bc GROUP 2 The leftover terms for each group are not identical and can not be used as the GCF for factoring as they are because: a(c – d) 4b(d – c) a(c – d)+ 4b(d – c) (d – c)(c – d) The signs for each term are opposite of each other. However these terms can be easily be made identical by removing a common factor of –1 out of one of them. -1(c – d)(d – c) = -1(4b)(c – d)4b(d – c) = -4b(c – d)4b(d – c) =

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ac - ad + 4bd – 4bc When you switch these terms around, the sign of the factor must change. = ac - ad GROUP 1 = a(c – d)+ 4b(d – c) - 4b(c – d) When reversing the order of terms of a binomial, it is important to understand the distinction of when to change the sign of the common factor. If the terms are added, there is no need to reverse the sign of the GCF, but if the terms are subtracted, there is a need to reverse the sign of the GCF. + 4bd – 4bc GROUP 2 = a(c – d) - 4b(c – d) The final step is to remove the common binomial factor. = (c – d) (a - 4b) (c + d)(d + c) = -1(c – d)(d – c) = (3yz + 7x)(7x + 3yz) = -1(3yz - 7x)(7x - 3yz) = When you switch these terms around,

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