Question

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x + x² – 2x – 1, g(x) = x + 1​

Answer 1

Answer:

(i) Apply factor theorem

x+1=0

So x=−1

2x

3

+x

2

−2x−1

Replace x by −1, we get

2(−1)

3

+(−1)

2

−2(−1)−1=−2+1+2−1=0

Reminder is 0 so that x+1 is a factor of 2x

3

+x

2

−2x−1

(ii) Apply factor theorem

x+2=0

So x=−2

x

3

+3x

2

+3x+1

Replace x by −2, we get

(−2)

3

+3(−2)

2

+3(−2)+1=−8+12−6+1=1

Reminder is 1 so that x+2 is not a factor of x

3

+3x

2

+3x+1

(iii) Apply factor theorem

x−3=0

So x=3

x

3

−4x

2

+x+6

Replace x by 3, we get

(3)

3

−4(3)

2

+(3)−1=27−36+3+6=0

Reminder is 0 so that x−3 is a factor of x

3

−4x

2

+x+6

Answer 2

""" ❤️ Answer ❤️ """

Apply factor theorem

x+

1=

0

So

x=

−1

2x

3

+

x 2

2x− 1

Replace x by −1 , we get

2(−1)

3

+ (−1)

2

− 2(−1)− 1= −2+ 1+ 2− 1= 0

2x

3

+

Reminder is 0 so that

x+

1

is a

x 2

2x−

1

(ii) Apply factor theorem

x+ 2= 0

So

x=

−2

x 3

3x

2

+ 3x+ 1

Replace x by −2 , we get

(−2)

3

+ 3(−2)

2

+ 3(−2)+ 1= −8+ 12− 6+ 1= 1

Reminder is 1 so that

x+

2

is not a

x 3

3x

2

+

3x+

1

(iii) Apply factor theorem

x− 3= 0

So

x=

3

x 3

4x

2

+ x+ 6

Replace x by 3 , we get

(3)

3

− 4(3)

2

+ (3)− 1= 27− 36+ 3+ 6= 0

Reminder is 0 so that

x−

3

is a

x 3

4x

2

+

x+

6