How To Factor Polynomials With 4 Terms Without Grouping

HOW TO FACTOR POLYNOMIALS WITH 4 TERMS WITHOUT GROUPING

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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)

Example 4 :

Factor the polynomial completely.

a) x3 − 7x2 + 10x

b) 3n7 − 75n5

c) 8m5 − 16m4 + 8m3

Solution :

a) x3 − 7x2 + 10x

= x(x2 − 7x + 10)

= x(x2 − 2x - 5x + 10)

= x[x(x - 2) - 5(x - 2)]

= x(x - 2)(x - 5)

b) 3n7 − 75n5

Factoring 3n5m, we get

= 3n5(n2 - 25)

= 3n5(n2 - 52)

= 3n5(n + 5)(n - 5)

c) 8m5 − 16m4 + 8m3

Factoring 8m3, we get

= 8m3 (m2− 2m + 1)

= 8m3 (m2− 1m - 1m + 1)

= 8m3 [m(m - 1) - 1(m - 1)]

= 8m3 (m - 1)(m - 1)

Example 5 :

Determine whether

(a) x − 2 is a factor of

f(x) = x2 + 2x − 4

and

(b) x + 5 is a factor of

f(x) = 3x4 + 15x3 − x2 + 25

Solution :

a) To check if x - 2 is a factor, let x - 2 = 0 and x = 2

While applying x = 2 in the polynomial f(x) when we get the remainder as 0, then it must be a solution and x - 2 is a factor.

f(2) = 22 + 2(2) − 4

= 4 + 4 - 4

= 4

x - 2 is not a factor of the polynomial.

b) x + 5 = 0

x = -5

f(x) = 3x4 + 15x3 − x2 + 25

f(-5) = 3(-5)4 + 15(-5)3 − (-5)2 + 25

= 3(625) + 15(-125) - 25 + 25

= 1875 - 1875 - 25 + 25

= 0

So 5 is the solution and x + 5 is a factor.

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