Find te values of a and b so that x^4+ x^3+ 8x^2+ax+b is divisible by x^2+1.

Asked by  | 13th May, 2012, 11:52: AM

Expert Answer:

Since x 2 + 1 divides x 4 + x 3 + 8x 2 + ax + b , so the quotient will be a polynomial of degree 2.

So, we can write

x 4 + x 3 + 8x 2 + ax + b = (x 2 + 1) (a 1 x 2 + b 1 x + c 1)

? x 4 + x 3 + 8x 2 + ax + b = a 1 x 4 + a 1 x 2 + b 1 x 3 + b 1 x + c 1 x 2 + c 1

? x 4 + x 3 + 8x 2 + ax + b = a 1 x 4 + b 1 x 3 + (a 1 + c 1) x 2 + b 1 x + c 1

Comparing the coefficient of x 4 on both sides, we get –

a 1 = 1

On comparing the coefficient of x 3, we get –

b 1 = 1

On comparing the coefficient of x 2, we get –

a 1 + c 1 = 8

? 1 + c 1 = 8

? c 1 = 7

On comparing the coefficient of x on both sides, we get –

a = b 1 = 1

? a = 1

On comparing the constants on both sides, we get –

b = c 1 = 7

? b = 7

Answered by  | 13th May, 2012, 04:05: PM

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