HOW TO FACTOR POLYNOMIALS WITH 4 TERMS WITHOUT GROUPING

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,

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,

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 3n^5m, 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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