Question

Q. 3. Find the values of a and b so that x4 + x3 + 8x2 + ax + b is divisible by x2 + 1.

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Answer 1

$\boxed{P(x)= x^{4}+ x^{3} +8x^{2} +ax+b}$

$\boxed{\sf{g(x) = x^{2}+1}}$

Given that :-

P(x) is completely divisible by g(x)

So,

According to $\sf{FACTOR\:THEOREM}$ :-

r(x) will be zero that is remainder will be 0.

$\rule{200}{3}$

Now,

After dividing we get

$\boxed{Quotient = x^{2} +x + 7}$

$\boxed{Remainder = (a-1)x+(b-7)}$

$\rule{200}{3}$

We know that remainder will be equal to zero.

So,

$\boxed{(a-1)x+(b-7) = 0}$

Or

$\boxed{(a-1)x+(b-7) = 0x + 0}$

On comparing we get :-

a - 1 =0

a =1

b-7 = 0

b = 7

Thus,

$\boxed{a=1}$

$\boxed{b=7}$

Answer 2

Answer: a =1 ; b = 7

Step-by-step explanation:

Given,

$p(x):\;\;x^4+x^3+8x^2+ax+b$

If x² + 1 is divisible  by p(x) then, putting x² + 1 = 0 or x² = -1 or x = √(-1) or x = i, We get remainder  = 0 (Here i is imaginary letter iota)

Also we know  that,

i⁴ = 1

i³ = -i

i² = -1

Now, Putting x = i

Remainder = i⁴ + i³ + 8i² + ai + b

0 = 1 - i - 8 + ai + b

0 = -7 + b + ai - i

0 = b - 7 + i(a - 1)

(b - 7) + i(a - 1) = 0 +0i

On comparing Real and Imaginary parts,

b - 7 = 0

b = 7

And,

a - 1 = 0

a = 1