For what value pf b will the polynomial p(x)=4x^3-3x^2+bx+6 have the same remainder when its divided by x-1 and by x-3?
Can someone help me get this started?
I think you can do synthetic division and equate the terms involving b. Or
p(x) = 4x^3-3x^2+bx+6
= 4x^2(x-1) + x^2 + bx + 6 (because -3x^2 = -4x^2 + x^2)
= 4x^2(x-1) + x(x-1) + (b+1)(x-1) + 6-b-1
Remainder when divided by x-1 is 5-b
p(x) = 4x^2(x-3) + 9x(x-3) + (b+27)(x-3) - 3b-81+6
So remainder = 3b-75
If 5 -b = 3b - 75 then 80 = 4b and b = 20 (if I've got my arithmetic right).
My method is simple. Sice the highest power in p(x) is x^3, consider
p(x) = (x-1)(ax^2+mx+c). It cannot be something like (x-1) * (zx^3+ax^2+mx+c) because there is no x^4 in p(x).
p(x) = ax^2(x-1) + mx(x-1) + c(x-1) = ax^3 - ax^2 + mx^2 - mx + cx - c
Equate coefficients. So
a = 4. (Coeff of x^3)
(-a + m) = -3 (Coeff of x^2) so -4 + m = -3 so m = 1.