Question

2. Use the Factor Theorem to determine whether g(x) is a factor of pox) in each of the
following cases: (i) p(x)=2x³ +x²-2x-1,g(x)=x+1
(ii)p(x)=x³+3x² + 3x + 1, g(x)=x+2
(iii) p(x)=x³-4x²+x+6, g(x)=x-3​

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