Question

find the value of A and B so that x to the power 4 + x cube + 8 x square + aX + b is divisible by X square + 1​

Answer 1

Step-by-step explanation:

let p(x) =

${x}^{4} + {x}^{3 } + 8 {x}^{2} + ax + b$

divisor=

${x}^{2} + 1$

Therefore,

p(-1)=

${ - 1}^{4 } + { - 1}^{3} + 8 \times \times { - 1}^{2} + a \times - 1 + b$

therefore,

1-1+8+a+b=p(-1)

8+a+b=p(-1)

but p(-1)=0 by factor theorem

therefore,

8+a+b=0

a+b=0-8

a+b=-8

therefore possible combinations are:-

1. 8,0
2. 7,1
3. 6,2
4. 5,3
5. 4,4

try these and find the answer

please mark this answer as brainliest