Question

3. Find the value of a and b when a x3 + bx2+x - 6 has (x+2) as a factor and leave
remainder 4 when divided by x - 2, find the value of a and b​

Answer 1

Answer:

Let p(x)=ax3+bx2+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                                 ...(i)

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                                        ...(ii)

Adding equation (i) and (ii), we get

(−2a+b)+(2a+b)=2+2

⇒2b=4⇒b=2

Putting b=2 in (i), we get

−2a+2=2

⇒−2a=0⇒a=0

Hence, a=0 and b=2

Answer 2

Second I’d givens 2x7