HOW TO FACTOR POLYNOMIALS WITH 4 TERMS WITHOUT GROUPING
Let P(x) be a polynomial with four terms.
To factor P(x) without grouping, substitute
x = -1, 1, -2, 2, -3, 3.......
P(-1) = 0 ----> (x + 1) is a factor of P(x)
P(1) = 0 ----> (x - 1) is a factor of P(x)
P(-2) = 0 ----> (x + 2) is a factor of P(x)
P(2) = 0 ----> (x - 2) is a factor of P(x)
If P(-1) ≠ 0, then (x + 1) is not a factor of P(x).
Then, try x = 1, x = -2, x = 2 and so on.
Once one of the linear factors of P(x) is found, the other factors can bound easily (the rest of the process has been explained in the following examples).
Factor the following polynomials without grouping :
Example 1 :
x3 - 2x2 - x + 2
Solution :
Let p(x) = x3 - 2x2 - x + 2.
Substitute x = -1.
p(-1) = (-1)3 - 2(-1)2 - (-1) + 2
= -1 - 2(1) + 1 + 2
= -1 - 2 + 1 + 2
= 0
Since P(-1) = 0, (x + 1) is a factor of P(x).
Since P(x) is a cubic polynomial, the other factor can be assumed as (x2 + ax + b).
Then,
(x + 1)((x2 + ax + b) = x3 - 2x2 - x + 2
Comparing the coefficients of x and constants,
|
b + a = -1 ----(1) |
b = 2 |
Substitute b = 2 in (1).
2 + a = -1
a = -3
x2 + ax + b = x2 - 3x + 2
Factors of (x2 - 3x + 2) are (x - 1) and (x - 2).
Therefore,
x3 - 2x2 - x + 2 = (x + 1)(x - 1)(x - 2)
Example 2 :
x3 + 4x2 + x - 6
Solution :
Let p(x) = x3 + 4x2 + x - 6.
Substitute x = -1.
p(1) = (-1)3 + 4(-1)2 + (-1) - 6
= -1 + 4(1) - 1 - 6
= -1 + 4 - 1 - 6
= -4 ≠ 0
Substitute x = 1.
p(1) = 13 + 4(1)2 + 1 - 6
= 1 + 4 + 1 - 6
= 0
Since P(1) = 0, (x - 1) is a factor of P(x).
Since P(x) is a cubic polynomial, the other factor can be assumed as (x2 + ax + b).
Then,
(x - 1)((x2 + ax + b) = x3 + 4x2 + x - 6
Comparing the coefficients of x and constants,
|
b - a = 1 ----(1) |
-b = -6 b = 6 |
Substitute b = 6 in (1).
6 - a = 1
-a = -5
a = 5
x2 + ax + b = x2 + 5x + 6
Factors of (x2 + 5x + 6) are (x + 2) and (x + 3).
Therefore,
x3 + 4x2 + x - 6 = (x - 1)(x + 2)(x + 3)
Example 3 :
x3 + 3x2 - 4x - 12
Solution :
Let p(x) = x3 + 3x2 - 4x - 12.
Substitute x = -1.
p(-1) = (-1)3 + 3(-1)2 - 4(-1) - 12
= -1 + 3(1) + 4 - 12
= -1 + 3 + 4 - 12
= -6 ≠ 0
Since P(-1) ≠ 0, (x + 1) is not a factor of P(x).
Substitute x = 1.
p(1) = 13 + 3(1)2 - 4(1) - 12
= 1 + 3 - 4 - 12
= -12 ≠ 0
Substitute x = -2.
p(-2) = (-2)3 + 3(-2)2 - 4(-2) - 12
= -8 + 3(4) + 8 - 12
= -8 + 12 + 8 - 12
= 0
Since P(-2) = 0, (x + 2) is a factor of P(x).
Since P(x) is a cubic polynomial, the other factor can be assumed as (x2 + ax + b).
Then,
(x + 2)((x2 + ax + b) = x3 + 3x2 - 4x - 12
Comparing the coefficients of x and constants,
b + 2a = -4 ----(1)2b = -12b = -6
Substitute b = -6 in (1).
-6 + 2a = -4
2a = 2
a = 1
x2 + ax + b = x2 + x - 6
Factors of (x2 + x - 6) are (x - 2) and (x + 3).
Therefore,
x3 + 3x2 - 4x - 12 = (x + 2)(x - 2)(x + 3)
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