FIND LINEAR FACTORS OF A POLYNOMIAL

Quadratic Polynomial :

The polynomial which is having highest exponent of 2 will have two linear factors.

To find factors, we can use factoring method, if it is not factorable we have to solve the quadratic equation by equating the polynomial to 0 and use quadratic formula

Once we get zeroes, convert them as factors.

Cubic Polynomial :

The polynomial which is having highest exponent of 3 will have three linear factors. The following methods will be helpful to find factors of cubic 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 = -b ± b2 - 4ac2az = -(-6) ± (-6)2 - 4(1)(16)2(1)z = 6 ± 36 - 642(1)z = 6 ± -282z = 6 ± -7(4) 2z = 6 ± 2i72z = 23 ±i 72z = 3 ± i7

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 = -b ± b2 - 4ac2ax = -2 ± 22 - 4(1)(-4)2(1)x = -2 ± 4 + 162(1)x = -2 ± 202x = -2 ± 5(4) 2x = -2 ± 252x = 2-1 ± 52x = -1 ± 5

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).

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