Question

find the value m and n such that x^4+x^3+8x^2+mx+n is divisible by x^2+ 1​

Answer 1

Answer:

Step-by-step explanation:

If f(x) =x^3-6x^2+mx-n is exactly divisible by x-1

So x-1=0

x=1

f(1)=1^3-6(1)^2+m×1-n=0

=1-6+m-n=0

=-5+m-n=0

=m=n+5 ... (1)

If f(x) =x^3-6x^2+mx-n is exactly divisible by x-2

So x-2=0

x=2

f(2)=2^3-6(2)^2+m×2-n=0

=8-24+2m-n=0

=-16+2m-n=0

=2m-n=16 ... (2)

From eq 1&2

2(n+5)-n=16[m=n+5]

2n+10-n=16

n=16-10

n=6

Putting the value of n in eq 1

m=6+5

m=11