To know how to factor quadratic expressions,
Example 1 :
Factor :
x2 + 13x + 30
Solution :
In the given quadratic expression, the coefficient of x2 is 1.
Decompose the constant term +30 into two factors such that the product of the two factors is equal to +30 and the addition of two factors is equal to the coefficient of x, that is +13.
Then, the two factors of +30 are
+3 and +10
Factor the given quadratic expression using +3 and +10.
x2 + 13x +30 = (x + 3)(x + 10)
Therefore, the factors of the given quadratic expression are
(x + 3) and (x + 10)
Example 2 :
Factor :
x2 + 22x + 21
Solution :
In the given quadratic expression, the coefficient of x2 is 1.
Decompose the constant term +21 into two factors such that the product of the two factors is equal to +21 and the addition of two factors is equal to the coefficient of x, that is 22.
Then, the two factors of -21 are
+1 and +21
Factor the given quadratic expression using +1 and +21.
x2 + 22x + 21 = (x + 1)(x + 21)
Therefore, the factors of the given quadratic expression are
(x + 1) and (x + 21)
Example 3 :
Factor :
x2 + 3x – 28
Solution :
In the given quadratic expression, the coefficient of x2 is 1.
Decompose the constant term -28 into two factors such that the product of the two factors is equal to -28 and the addition of two factors is equal to the coefficient of x, that is +3.
Then, the two factors of -28 are
-4 and +7
Factor the given quadratic expression using -4 and +7.
x2 + 3x – 28 = (x - 4)(x + 7)
Therefore, the factors of the given quadratic expression are
(x - 4) and (x + 7)
Example 4 :
Factor :
x2 – x – 30
Solution :
In the given quadratic expression, the coefficient of x2 is 1.
Decompose the constant term -30 into two factors such that the product of the two factors is equal to -30 and the addition of two factors is equal to the coefficient of x, that is -1.
Then, the two factors of -30 are
-6 and +5
Factor the given quadratic expression using -6 and +5.
x2 – x – 30 = (x - 6)(x + 5)
Therefore, the factors of the given quadratic expression are
(x - 6) and (x + 5)
Example 5 :
Factor :
x2 – 2x – 24
Solution :
= x2 – 2x – 24
= x2 – 6x + 4x – =24
= x(x - 6) + 4(x - 6)
= (x - 6)(x + 4)
Example 6 :
Factor :
3x2 + 7x – 6
Solution :
= 3x2 + 7x – 6
= 3x2 + 9x - 2x – 6
= 3x(x + 3) - 2(x + 3)
= (3x - 2)(x + 3)
Example 7 :
Factor :
-x2 + 5x + 84
Solution :
= -x2 + 5x + 84
= -x2 + 12x - 7x + 84
= -x(x - 12) - 7(x - 12)
= (-x - 7)(x - 12)
Example 8 :
Factor :
-6x2 + x + 2
Solution :
= -6x2 + x + 2
= -6x2 + 4x - 3x + 2
= -2x(3x - 2) - 1(3x - 2)
= (-2x - 1)(3x - 2)
Example 9 :
3x2 - 48
Which of the following is equivalent to the expression above.
a) 3(x - 4)(x + 4) b) 3(x - 4)2 c) (3x - 4)(3x + 4)
d) (3x + 4)(x - 4)
Solution :
= 3x2 - 48
Factoring 3, we get
= 3(x2 - 16)
= 3(x2 - 42)
Using algebraic identity a2 - b2, we get (a + b) (a - b)
= 3(x + 4)(x - 4)
So, option a is correct.
Example 10 :
If x + y = 10 and x - y = 4, what is the value of x2 - y2 ?
a) 20 b) 24 c) 36 d) 40
Solution :
x2 - y2 = (x + y)(x - y)
Given values are x + y = 10 and x - y = 4
x2 - y2 = 10(4)
= 40
so, option d is correct.
Example 11 :
6x2 + 7x - 24 = 0
If r and s are two solutions of the equation above and r > s, which of the following is the value of r - s ?
a) 7/6 b) 16/3 c) 25/6 d) 20/3
Solution :
6x2 + 7x - 24 = 0
Solving this quadratic equation using factoring method,
6x2 + 16x - 9x - 24 = 0
2x(3x + 8) - 3(3x + 8) = 0
(2x - 3)(3x + 8) = 0
Equating each factor to 0, we get
2x - 3 = 0 2x = 3 x = 3/2 |
3x + 8 = 0 3x = - 8 x = -8/3 |
So, the solutions are 3/2 and -8/3
r = 3/2 > s = -8/3
r - s = 3/2 - (-8/3)
= 3/2 + (8/3)
= (9 + 16)/6
= 25/6
So, option c is correct.
Example 12 :
x2 - 3x = 28
If r and s are two solutions of the equation above, which of the following is the value of r + s ?
a) -3 b) 3 c) 6 d) 9
Solution :
x2 - 3x = 28
x2 - 3x - 28 = 0
x2 + 7x - 4x - 28 = 0
x(x + 7) - 4(x + 7) = 0
(x - 4)(x + 7) = 0
Equating each factor to 0, we get
x - 4 = 0 and x + 7 = 0
x = 4 and x = 7
r =4 and s = 7
r + s = 4 + 7
= 11
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