MTH 030 -- Elementary Algebra -- Exercise Solutions
Section:
5.2

30)  Factor x2 + 7x + 10
       Solution:

x2 + 7x + 10 = (x + 2)(x + 5)
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Find two numbers whose product is 10 and whose sum is 7. They will be the constant terms in the linear factors. So the constant terms will be 2 and 5.

32)  Factor n2 - 7n + 10
       Solution:   n2 - 7n + 10 = (n - 2)(n - 5)

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Find two numbers whose product is 10 and whose sum is - 7. They will be the constant terms in the linear factors. Because 10 is positive, the two numbers must have the same sign and since - 7 is negative that sign is minus. So the constant terms will be - 2 and - 5.

34)  Factor b2 + 6b - 7
       Solution:   b2 + 6b - 7 = (b + 7)(b - 1)


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Find a positive number and a negative number whose product is -7 and whose sum is 6. They will be the constant terms in the linear factors. Because - 7 is negative, the two numbers must have different signs and since +6 is positive the larger of the two is positive. So the constant terms will be 7 and - 1.

36)  Factor t2 - 5t - 50
       Solution:  t2 - 5t - 50 = (t - 10)(t + 5)


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Find a positive number and a negative number whose product is - 50 and whose sum is - 5. They will be the constant terms in the linear factors. Because - 50 is negative, the two numbers must have different signs and since - 5 is negative the larger of the two is negative. So the constant terms will be -10 and 5.

38)  Factor r2 - 9r - 12
       Solution:  This is a prime polynomial

Find a positive number and a negative number whose product is - 12and whose sum is - 9. They will be the constant terms in the linear factors. Because - 12 is negative, the two numbers must have different signs and since - 9 is negative the larger of the two is negative. No pair of divisors of 12 can fit these conditions so the trinomial is prime.

40)  Factor v2 + 9v + 15
       Solution:  This is a prime polynomial

Find two numbers whose product is 15 and whose sum is 9. They will be the constant terms in the linear factors. No pair of divisors of 15 can fit these conditions so the trinomial is prime.

42)  Factor y2 + 8y + 12
       Solution:  y2 + 8y + 12 = (y + 6)(y + 2)

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Find two numbers whose product is 12 and whose sum is 8. They will be the constant terms in the linear factors. So the constant terms will be 2 and 6.

44)  Factor m2 + 3m - 10
       Solution:  m2 + 3m - 10 = (m + 5)(m - 2)


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Find a positive number and a negative number whose product is - 10 and whose sum is 3. They will be the constant terms in the linear factors. Because - 10 is negative, the two numbers must have different signs and since + 3 is positive the larger of the two is positive. So the constant terms will be 5 and - 2.

46)  Factor u2 + u - 42
       Solution:  u2 + u - 42 = (u + 7)(u - 6)


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Find a positive number and a negative number whose product is - 42 and whose sum is 1. They will be the constant terms in the linear factors. Because - 42 is negative, the two numbers must have different signs and since + 1 is positive the larger of the two is positive. So the constant terms will be 7 and - 6.

48)  Factor a2 + 10ab + 9b2
       Solution:  a2 + 10ab + 9b2 = (a + 9b)(a + b)

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Find two numbers whose product is 9 and whose sum is 10. They will be the coefficients of b in the linear factors. So the second terms will be 9b and b.

50)  Factor m2 - mn - 12n2
       Solution:  m2 - mn - 12n2 = (m - 4n)(m + 3n)


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Find a positive number and a negative number whose product is - 12 and whose sum is -1. They will be the coefficients of n in the linear factors. Because - 12 is negative, the two numbers must have different signs and since - 1 is negative the larger of the two is negative. So the second terms will be - 4n and 3n.

52)  Factor p2 + pq - 6q2
       Solution:  p2 + pq - 6q2 = (p + 3q)( p - 2q)


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Find a positive number and a negative number whose product is - 6 and whose sum is 1. They will be the coefficients of q in the linear factors. Because - 6 is negative, the two numbers must have different signs and since + 1 is positive the larger of the two is positive. So the second terms will be 3q and - 2q.

54)  Factor m2 + 3mn - 20n2
       Solution:  This is a prime polynomial

Find a positive number and a negative number whose product is - 20 and whose sum is 3. They will be the coefficients of q in the linear factors. No pair of divisors of 15 can fit these conditions so the trinomial is prime.

56)  Factor -x2 + 9x - 20
       Solution:  -x2 + 9x - 20 = (-1)[x2 - 9x + 20]
       =(-1)[(x- 4)(x- 5)] = (-1)(x- 4)(x- 5)
       = - (x- 4)(x- 5)

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58)  Factor - t2 - t + 30
       Solution: - t2 - t + 30 = (-1)[t2 + t - 30]
        = (-1)[(t + 6)(t - 5)] = (-1)(t + 6)(t - 5)
       
= - (t + 6)(t - 5)

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60)  Factor - r2 + 14r - 45
       Solution:  - r2 + 14r - 45 = (-1)[r2 - 14r + 45]
       = (-1)[(r - 9)(r - 5)] = (-1)(r - 9)(r - 5)
       = - (r - 9)(r - 5)

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62)  Factor - a2 - 6ab - 5b2
       Solution:  - a2 - 6ab - 5b2 = (-1)[a2 + 6ab + 5b2]
       = (-1)[(a +5b)(a + b)] = (-1)(a +5b)(a + b)
       = - (a +5b)(a + b)

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64)  Factor - x2 - 10xy + 11y2
       Solution:  - x2 - 10xy + 11y2 = (-1)[x2 +10xy - 11y2
        = (-1)[(x + 11y)(x - y)] = (-1)(x + 11y)(x - y)
        = - (x + 11y)(x - y)

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66)  Factor y2 + 5 + 6y
       Solution:  y2 + 5 + 6y = y2 + 6y + 5 
       = (y + 5)(y + 1)



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68)  Factor x2 - 13 - 12x
       Solution:  x2 - 13 - 12x = x2 - 12x - 13
       = (x + 1)(x - 13)

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70)  Factor u2 - 3 + 2u
       Solution:  u2 - 3 + 2u = u2 + 2u - 3
        = (u + 3)(u - 1)

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72)  Factor a2 + 5b2 + 6ab
       Solution:  a2 + 5b2 + 6ab = a2+ 6ab + 5b2
       = (a + 5b)(a + b)

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74)  Factor -13yz + y2 - 14z2
       Solution:  -13yz + y2 - 14z2 = y2- 13yz - 14z2
         = (y - 14z)(y + z)

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76)  Factor 3y2 - 21y + 18
       Solution:  3y2 - 21y + 18 = 3[y2 - 7y + 6]
       = 3[(y - 6)(y - 1)] = 3(y - 6)(y - 1)

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78)  Factor - 2b2 + 20b - 18
       Solution:  - 2b2 + 20b - 18 = (-2)[b2 - 10b + 9]
       = (-2)[(b - 9)(b - 1)] = (-2)(b - 9)(b - 1)

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80)  Factor 5m2 + 45m - 50
       Solution:  5m2 + 45m - 50 = 5[m2 + 9m - 10]
       = 5(m - 1)(m + 10)

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82)  Factor 48xy + 6xy2 + 96x
       Solution:  48xy + 6xy2 + 96x = 6xy2 + 48xy + 96x
       = 6x[y2 + 8y + 16] = 6x[(y + 4)(y + 4)]
       = 6x(y+ 4)2


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84)  Factor 3x2y3 + 3x3y2 - 6xy4
       Solution:  3x2y3 + 3x3y2 - 6xy4 = 3xy2[xy + x2 - 2y2]
        = 3xy2[x2+ xy - 2y2] = 3xy2[(x + 2y)(x - y)]
        = 3xy2(x + 2y)(x - y)

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