Question

Al Ax^3 + bx^2 + x - 6 has x +2
as a factor and leave remainder 4 when divided by x-2​

Answer 1

Let p(x)=ax 3 +bx 2+x−6

Since (x+2) is a factor of p(x), then by Factor theorem p(−2)=0

⇒a(−2)

3

+b(−2)

2

+(−2)−6=0

⇒−8a+4b−8=0

⇒−2a+b=2

Also when p(x) is divided by (x-2) the remainder is 4, therefore by Remainder theorem p(2)=4

⇒a(2)

3

+b(2)

2

+2−6=4

⇒8a+4b+2−6=4

⇒8a+4b=8

⇒2a+b=2

Adding equation

(−2a+b)+(2a+b)=2+2

⇒2b=4⇒b=2

Putting b=2 in (i), we get

−2a+2=2

⇒−2a=0⇒a=0

,a=0 and b=2

I HOPE THIS IS HELPFUL FOR YOU ☺️