When x^4 - 2x^3 - 4b^2x^2 - 40 bx - 56 is divided by x+2b, the remainder is 200.
Show that b^3 + 5b^2 - 16 = 0
Find the possible values of b
Yes, I'd need help with part 2. Thanks .
If it's O level I think you have to guess 1 of the factor which is an integer.
In this case, (b+4) is a factor
Then b³+5b²-16 = (b+4)(b²+b-4)
= 0
Then solve the quadratic (b²+b-4=0) and linear equation(b+4=0) to get the values.