Puzzling Equation?

Question

Find the Values of a and b so that x^4 +x^3 + 8x^2 +ax + b is divisible by x^2 + 1

Answer 1

The polynomial x^4 +x^3 + 8x^2 +ax + b divided by x^2 + 1 is x^2 + x + 7 with a remainder of (a-1)x + (b-7). So for x^4 +x^3 + 8x^2 +ax + b to be divisible by x^2 + 1, we need (a-1)=0 and (b-7)=0, so a=1 and b=7.

Answer 2

Let p(x) = x^4 + x^3 + 8x^2 + ax + b

One thing to note is that x^2 + 1 factors into (x - i) (x + i). This means that in order for p(x) to be divisible by (x^2 + 1),

p(i) = 0 and p(-i) = 0. However, by the formula
p(i) = i^4 + i^3 + 8i^2 + ai + b

Note that i^2 = (-1); therefore

p(i) = (i^2)(i^2) + i(i^2) + 8(-1) + ai + b
p(i) = (-1)(-1) + i(-1) - 8 + ai + b
p(i) = 1 - i - 8 + ai + b
p(i) = (b - 7) + ai - i
p(i) = (b - 7) + (a - 1)i

Since p(i) = 0, it follows that

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

0 = 0 + 0i, so

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

So we can equate complex numbers component-wise. This means

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

Therefore, the values of a and b that make p(x) divisible by
x^2 + 1 is a = 1, b = 7. The function itself is then

p(x) = x^4 + x^3 + 8x^2 + x + 7

As a side note, if we perform long division with x^2 + 1, we will obtain

p(x) = (x^2 + 1) (x^2 + x + 7)

Answer 3

We know that x^2+1 is a factor of the equation, so it can be rewritten as: (x^2+1)(x^2+cx+d) for some c and d. Equating this with the original equation gives us:

(x^2+1)(x^2+cx+d)=x^4 +x^3 + 8x^2 +ax + b
x^4+cx^3+dx^2+x^2+cx+d = x^4 +x^3 + 8x^2 +ax + b
x^4 + cx^3+(d+1)x^2+cx +d = x^4 +x^3 + 8x^2 +ax + b

equating the multipliers of each term we get:
c=1 (from x^3 term)
d+1 = 8; d=7 (from x^2)
a=c=1 (from x)
d=b=7 (constant)

Answer 4

(x^2 + 1)n = x^4 +x^3 + 8x^2 +ax + b
to get x^4 term n should contain term x^2
to get x^3 term n should contain x term
so n = ( x^2 + x + m)
where m has to be founded
ax + b will be taken care of
(x^2 + 1)( x^2 + x + m)=x^4 +x^3 + 8x^2 +ax + b
x^4 + x^3 + mx^2 + x^2 + x + m = x^4 +x^3 + 8x^2 +ax + b
mx^2 + x^2 = 8x^2 ( coeficient of x^2 on lhs has to be equal to coefficient of x^2 on rhs )
therefore (m+1) = 8
therefore m = 7
similarly
1 = a ( coeficient of x on lhs has to be equal to coefficient of x on rhs )
m = b
but m = 7
therefore b = 7
therefore a = 1 and b = 7

Answer 5

As ashish noted x=I or x=-I, thus plug any into source equation and see that a=1 and b=7; no sweat!

Answer 6

As ashish noted x=I or x=-I, thus plug any into source equation and see that a=1 and b=7; no sweat!

Answer 7

from x^2+1
we got x=+i,-i


so a=-1,b=7