MTH
030 -- Elementary Algebra -- Exercise Solutions
Section:
5.3
27)
Factor 2x^{2} - 3x + 1 Solution: 2x^{2} - 3x + 1 = (2x - 1)(x - 1) justify by multiplying |
The constant term is positive so the two factors will have the same sign. Because the middle term is negative, the sign in both factors will be -. |
28)
Factor 2y^{2} - 7y + 3 |
The constant term is positive so the two factors will have the same sign. Because the middle term is negative, the sign in both factors will be -. |
30)
Factor 2b^{2} + 7b + 6 |
The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
32)
Factor 4t^{2} - 4t + 1 Solution: 4t^{2} - 4t + 1 = (2t - 1)(2t - 1) = (2t - 1)^{2} justify by multiplying |
Both the leading term and the constant term are perfect squares and twice the cross product is the middle term. Therefore this trinomial factors as the square of a difference. |
34)
Factor 4x^{2} + 8x + 3 Solution: 4x^{2} + 8x + 3 = (2x + 3)(2x + 1) justify by multiplying |
The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
36)
Factor 4z^{2} - 9z + 2 |
The constant term is positive so the two factors will have the same sign. Because the middle term is negative, the sign in both factors will be -. |
38)
Factor 8u^{2} - 2u - 15 |
The constant term is negative so one factor will have a minus sign the other will have a plus sign. |
40)
Factor 12y^{2} - y - 1 |
The constant term is negative so one factor will have a minus sign the other will have a plus sign. |
42)
Factor 10u^{2} - 13u - 6 justify by multiplying |
Click Here to see all the possible factors |
44)
Factor 6m^{2} + 19m + 3 |
The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
46)
Factor 10x^{2} + 21x - 10 Solution: 10x^{2} + 21x - 10 = (2x + 5)(5x - 2) justify by multiplying |
The constant term is negative so one factor will have a minus sign the other will have a plus sign. |
48) Factor - 16y^{2} - 10y - 1 Solution: - 16y^{2} - 10y - 1 =(-1)(16y^{2} + 10y + 1) = (-1)(8y + 1)(2y + 1) = - (8y + 1)(2y + 1) justify by multiplying |
As
a first step factor -1 from the expression. |
50) Factor - 16x^{2} - 16x - 3 Solution: - 16x^{2} - 16x - 3 = (-1)(16x^{2} + 16x + 3) = (-1)(4x + 1)(4x + 3) = -(4x + 1)(4x + 3) justify by multiplying |
As
a first step factor -1 from the expression. Then observe the following about the trinomial. The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
52) Factor 2b^{2} - 5bc + 2c^{2} Solution: 2b^{2} - 5bc + 2c^{2} = (2b - c)(b - 2c) justify by multiplying |
The constant term is positive so the two factors will have the same sign. Because the middle term is negative, the sign in both factors will be -. |
52 Altrernate) Factor 2c^{2} - 5bc + 2b^{2} Solution: 2c^{2} - 5bc + 2b^{2} = (2c - b)(c - 2b) justify by multiplying |
The constant term is positive so the two factors will have the same sign. Because the middle term is negative, the sign in both factors will be -. |
54) Factor 3m^{2} + 5mn + 2n^{2} Solution: 3m^{2} + 5mn + 2n^{2} = (3m + 2n)(m + n) justify by multiplying |
The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
54 Alternate) Factor 2n^{2} + 5mn + 3m^{2} Solution: 2n^{2} + 5mn + 3m^{2} = (2n + 3m)(n + m) justify by multiplying |
The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
56) Factor 4b^{2} + 15bc - 4c^{2} Solution: 4b^{2} + 15bc - 4c^{2} = (4b - c)(b + 4c) justify by multiplying |
The constant term is negative so one factor will have a minus sign the other will have a plus sign. |
58) Factor 12x^{2} + 5xy - 3y^{2} Solution: 12x^{2} + 5xy - 3y^{2} = (4x + 3y)(3x - y) justify by multiplying |
The constant term is negative so one factor will have a minus sign the other will have a plus sign |
60) Factor - 14 + 3a^{2} - a Solution: - 14 + 3a^{2} - a = 3a^{2} - a - 14 = (3a - 7)(a + 2) justify by multiplying |
As
a first step write the trinomial in descending powers of the variable. Then observe the following about the trinomial. The constant term is negative so one factor will have a minus sign the other will have a plus sign |
62) Factor 16 - 40a + 25a^{2} Solution: 16 - 40a + 25a^{2} = 25a^{2}- 40a + 16 = (5a - 4)(5a - 4) = (5a - 4)^{2} justify by multiplying |
As
a first step write the trinomial in descending powers of the variable. Then observe the following about the trinomial. The first and last terms are perfect squares and the middle term is twice the cross product . Therefore the trinomial is a perfect square. Since the middle term is negative, the trinomial is the square of a difference. |
64) Factor 12t^{2} - 1 - 4t Solution: 12t^{2} - 1 - 4t = 12t^{2} - 4t - 1 = (6t + 1)(2t - 1) justify by multiplying |
As
a first step write the trinomial in descending powers of the variable. Then observe the following about the trinomial. The constant term is negative so one factor will have a minus sign the other will have a plus s |
66) Factor 25 + 2u^{2} + 3u |
As
a first step write the trinomial in descending powers of the variable. Then observe the following about the trinomial. The constant term is positive so the two factors will have the same sign. Because the middle term is positive, the sign in both factors will be +. |
68) Factor 11 uv + 3u^{2} + 6v^{2} |
70) Factor
12m^{2} - 1mn + 2n^{2} Solution: 12m^{2} - 1mn + 2n^{2} = (3m - 2n)(4m - n) justify by multiplying |
72) Factor 9x^{2} + 21x - 18 Solution: 9x^{2} + 21x - 18 = 3(3x^{2} + 7x - 6) = 3(3x - 2)(x + 3) justify by multiplying |
74) Factor - 2xy^{2} - 8xy + 24x Solution: - 2xy^{2} - 8xy + 24x = = - 2x(y^{2} + 4y - 12) = - 2x(y + 6)(y - 2) justify by multiplying |