FIND LINEAR FACTORS OF A POLYNOMIAL

Find the linear factors of :

Problem 1 :

2x² - 7x - 15

Solution :

2x² - 7x - 15

2x² - 10x + 3x - 15

2x(x - 5) + 3(x - 5)

(2x + 3) (x - 5)

So, the linear factors are (2x + 3) (x - 5).

Problem 2 :

z² - 6z + 16

Solution :

z² - 6z + 16

By using quadratic formula.

a = 1, b = -6 and c = 16

z = 3 + i√7 and z = 3 - i√7

(z - 3 + i√7) (z - 3 - i√7)

So, the linear factors are (z - 3 + i√7) (z - 3 - i√7).

Problem 3 :

x³ + 2x² - 4x

Solution :

x³ + 2x² - 4x

x(x² + 2x - 4)

By using quadratic formula.

a = 1, b = 2 and c = -4

x = (-1 + √5) and x = (-1 -  √5) 

x + 1 - √5 = 0 and x + 1 + √5 = 0

x(x + 1 + √5) (x + 1 - √5)

So, the linear factors are x(x + 1 + √5) (x + 1 - √5). 

Problem 4 :

6z³ - z² - 2z

Solution :

6z³ - z² - 2z

z(6z² - z - 2)

z(6z² + 3z - 4z - 2)

z(3z (2z + 1) - 2 (2z + 1))

z(3z - 2) (2z + 1)

So, the linear factors are z(3z - 2) (2z + 1). 

Problem 5 :

z⁴ - 6z² + 5

Solution :

z⁴ - 6z² + 5

z⁴ - z² - 5z² + 5 = 0

z²(z² - 1) - 5(z² - 1) = 0

(z² - 5) (z² - 1) = 0

z² - 5 = 0

z² = 5

Squaring on each sides.

z = ±√5

(z + √5) (z - √5)

z² - 1 = 0

z² = 1

Squaring on each sides.

z = ±√1

z =  ±1

(z + 1) (z - 1)

So, the linear factors are (z + 1) (z - 1) (z + √5) (z - √5).

Problem 6 :

z⁴ - z² - 2

Solution :

z⁴ - z² - 2

z⁴ + z² - 2z² - 2 = 0

z²(z² + 1)- 2(z² + 1) = 0

(z² - 2) (z² + 1) = 0

z² - 2 = 0

z² = 2

Squaring on each sides.

z =  ±√2

(z +√2) (z - √2)

z² + 1 = 0

z² = -1

Squaring on each sides.

z = ±i

(z + i) (z - i)

So, the linear factors are (z + i) (z - i) (z +√2) (z - √2).