Question

Find values of a and b so that x^4 + x^3 + 8x^2+ ax + bis divisible by x^2+ 13

Answer 1

If x²+13 is a factor of x^4+x^3+8x^2+ax+b

So,when the value of x from x²+13 is put into x^4+x^3+8x^2+ax+b will be equal to 0.

x²+13=0

x²=(-13)

x=√-13

Now,

(√-13)^4+(√-13)^3+8(√-13)^2 +a(√-13) +b =0

=169+(13√-13)-104+(a√-13)+b=0

=65+(√-13)(13+a)+b=0

=13+a+b=(-65/√-13)

=13+a+b=(-65/√-13×√-13/√-13)

=a+b=(-65√-13/-13)

=a+b=5√-13