Factoring Quadratic Trinomials in the form ax2 + bx + c, where a = 1.
Factoring Quadratic Trinomials in the form ax2 + bx + c, where a ≠ 1.
Factor each quadratic trinomial completely.
Problem 1 :
x2 + 7x + 12
Solution :
= x2 + 7x + 12
= x2 + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Problem 2 :
x2 + 5x - 14
Solution :
= x2 + 5x - 14
= x2 + 7x - 2x - 14
= x(x + 7) - 2(x + 7)
= (x + 7)(x - 2)
Problem 3 :
x2 - 11x + 30
Solution :
= x2 - 11x + 30
= x2 - 6x - 5x + 30
= x(x - 6) - 5(x - 6)
= (x - 6)(x - 5)
Problem 4 :
x2 - 5x - 24
Solution :
= x2 - 5x - 24
= x2 - 8x + 3x - 24
= x(x - 8) + 3(x - 8)
= (x - 8)(x + 3)
Problem 5 :
x2 - 7x - 18
Solution :
= x2 - 7x - 18
= x2 - 9x + 2x - 18
= x(x - 9) + 2(x - 9)
= (x - 9)(x + 2)
Problem 6 :
x2 + x - 20
Solution :
= x2 + x - 20
= x2 + 5x - 4x - 20
= x(x + 5) - 4(x + 5)
= (x + 5)(x - 4)
Problem 7 :
x2 - 3x - 40
Solution :
= x2 - 3x - 40
= x2 - 8x + 5x - 40
= x(x - 8) + 5(x - 8)
= (x - 8)(x + 5)
Problem 8 :
x2 + 3x - 28
Solution :
= x2 + 3x - 28
= x2 + 7x - 4x - 28
= x(x + 7) - 4(x + 7)
= (x + 7)(x - 4)
Problem 9 :
3x2 + 15x - 150
Solution :
= 3x2 + 15x - 150
= 3(x2 + 5x - 50)
= 3(x2 + 10x - 5x - 50)
= 3[x(x + 10) - 5(x + 10)]
= 3(x + 10)(x - 5)
Problem 10 :
2x2 + x - 3
Solution :
= 2x2 + x - 3
= 2x2 + 3x - 2x - 3
= x(2x + 3) - 1(2x + 3)
= (2x + 3)(x - 1)
Problem 11 :
6x2 + 17x + 5
Solution :
= 6x2 + 17x + 5
= 6x2 + 15x + 2x + 5
= 3x(2x + 5) + 1(2x + 5)
= (2x + 5)(3x + 1)
Problem 12 :
2x2 + 7x + 3
Solution :
= 2x2 + 7x + 3
= 2x2 + 6x + 1x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)
Problem 13 :
9x2 - 15x + 4
Solution :
= 9x2 - 15x + 4
= 9x2 - 12x - 3x + 4
= 3x(3x - 4) - 1(3x - 4)
= (3x - 4)(3x - 1)
Problem 14 :
10x2 + 21x + 9
Solution :
= 10x2 + 21x + 9
= 10x2 - 15x - 6x + 9
= 5x(2x - 3) - 3(2x - 3)
= (2x - 3)(5x - 3)
Problem 15 :
12x2 - 38x + 20
Solution :
= 12x2 - 38x + 20
= 2(6x2 - 19x + 10)
= 2(6x2 - 15x - 4x + 10)
= 2[3x(2x - 5) - 2(2x - 5)]
= 2(2x - 5)(3x - 2)
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