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Factor by grouping: a2 + a - 20. We will express the middle term, a, of the trinomial as the difference of two carefully chosen terms. We want to produce an equivalent four-term polynomial that can be factored by grouping. Since a? + a – 20 = a? + la - 20, we identify b as 1 and the key number c as -20. We must find two integers whose product is -20 and whose sum is 1. Since the integers must have a negative product, their signs must be different. Key number = -20 b = 1 Factors of -20 Sum of the factors of -20 1(-20) = -20 1 + (-20) = -19 2(-10) = -20 2 + (-10) = -8 4(-5) = -20 4 + (-5) = -1 5(-4) = -20 5 + (-4) = 1 10(-2) = -20 10 + (-2) = 8 20(-1) = -20 20 + (-1) = 19 The fourth row of the table contains the correct pair of integers 5 and -4, whose product is -20 and whose sum is 1. They serve as the coefficients of 5a and -4a, the two terms that we use to represent the middle term, a, of the trinomial. a2 + a - 20 = a2 + 5a - 4a - Express the middle term, a, as 5a – 4a. - a (a + 0) - «(a +0) - (a + O)(- -0) Factor a out of a2 + 5a and -4 out of -4a – 20. Factor out a + 5. Check the factorization by multiplying. Factor: m2 + m – 12.

Factor by grouping: a2 + a - 20. We will express the middle term, a, of the trinomial as the difference of two carefully chosen terms. We want to produce an equivalent four-term polynomial that can be factored by grouping. Since a? + a – 20 = a? + la - 20, we identify b as 1 and the key number c as -20. We must find two integers whose product is -20 and whose sum is 1. Since the integers must have a negative product, their signs must be different. Key number = -20 b = 1 Factors of -20 Sum of the factors of -20 1(-20) = -20 1 + (-20) = -19 2(-10) = -20 2 + (-10) = -8 4(-5) = -20 4 + (-5) = -1 5(-4) = -20 5 + (-4) = 1 10(-2) = -20 10 + (-2) = 8 20(-1) = -20 20 + (-1) = 19 The fourth row of the table contains the correct pair of integers 5 and -4, whose product is -20 and whose sum is 1. They serve as the coefficients of 5a and -4a, the two terms that we use to represent the middle term, a, of the trinomial. a2 + a - 20 = a2 + 5a - 4a - Express the middle term, a, as 5a – 4a. - a (a + 0) - «(a +0) - (a + O)(- -0) Factor a out of a2 + 5a and -4 out of -4a – 20. Factor out a + 5. Check the factorization by multiplying. Factor: m2 + m – 12.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter8: Polynomials
Section8.6: Factoring Quadratic Trinomials
Problem 59PFA
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I need help with all of them plz.

Factor by grouping: a2 + a - 20. We will express the middle term, a, of the trinomial as the difference of two carefully chosen terms. We want to produce an equivalent four-term polynomial that can be factored by grouping. Since a + a – integers must have a negative product, their signs must be different. 20 = a2 + la – 20, we identify b as 1 and the key number c as -20. We must find two integers whose product is -20 and whose sum is 1. Since the Key number = -20 b = 1 Factors of –20 Sum of the factors of –20 1(-20) = -20 1 + (-20) = -19 2(-10) = -20 2 + (-10) = –8 4(-5) = -20 4 + (-5) = -1 5(-4) = -20 5 + (-4) = 1 10(-2) = -20 10 + (-2) = 8 20(-1) = -20 20 + (-1) = 19 The fourth row of the table contains the correct pair of integers 5 and -4, whose product is -20 and whose sum is 1. They serve as the coefficients of 5a and -4a, the two terms that we use to represent the middle term, a, of the trinomial. a2 + a - 20 = a2 + 5a 4а — 20 Express the middle term, a, as 5a ·4a. | ) - (a +D)(- -O) = a a + 4 a + Factor a out of a² + 5a and -4 out of -4a – 20. а — Factor out a + 5. Check the factorization by multiplying. Factor: m² + m – - 12.
Transcribed Image Text:Factor by grouping: a2 + a - 20. We will express the middle term, a, of the trinomial as the difference of two carefully chosen terms. We want to produce an equivalent four-term polynomial that can be factored by grouping. Since a + a – integers must have a negative product, their signs must be different. 20 = a2 + la – 20, we identify b as 1 and the key number c as -20. We must find two integers whose product is -20 and whose sum is 1. Since the Key number = -20 b = 1 Factors of –20 Sum of the factors of –20 1(-20) = -20 1 + (-20) = -19 2(-10) = -20 2 + (-10) = –8 4(-5) = -20 4 + (-5) = -1 5(-4) = -20 5 + (-4) = 1 10(-2) = -20 10 + (-2) = 8 20(-1) = -20 20 + (-1) = 19 The fourth row of the table contains the correct pair of integers 5 and -4, whose product is -20 and whose sum is 1. They serve as the coefficients of 5a and -4a, the two terms that we use to represent the middle term, a, of the trinomial. a2 + a - 20 = a2 + 5a 4а — 20 Express the middle term, a, as 5a ·4a. | ) - (a +D)(- -O) = a a + 4 a + Factor a out of a² + 5a and -4 out of -4a – 20. а — Factor out a + 5. Check the factorization by multiplying. Factor: m² + m – - 12.
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