Determine where p,q are constants, given that, on division by , the remainder is .
I found
I don't think I'm on the right track, but I don't know how else to work it out.
Thanks!
Good start. I am not sure if the following is the simplest way, but it seems to lead to finding p and q, at least in theory.
So, we have for some function . Now, looking at degrees of polynomials, is a quadratic polynomial, let's say .
Next, we look at both sides' coefficients of , , as well as the free terms. Equating the leading coefficients, we see that . Similarly, equating the constant terms (or putting x = 0), we get that . Finally, equating the coefficients at , we get that . Therefore, .
Next, one of the roots of is -1/2. Substituting into the original equation, we get , or . Substitute the expression for above; this gives (after lots of simplifications) .
Finally, combining this with , one can find and . In my calculations, this boils down to a linear (not even a quadratic) equation.
Disclaimer: my calculations can be wrong.
Given problem is
Since (2x^2 - 8x +8) is divisible by (x-2)^2
Compare the coefficients of x^3....> 3p^2 = a...(1)
Compare the coefficients of x^2...> 6pq - 11p^2 = -4a + b....(2)
Compare the coefficients of x....> 3q^2 - 22q = 4a - 4b - 3....(3)
Compare the constants....> -11q^2 = 4b - 39...(4)
Adding them you get
-8(p+q)^2 = a + b - 42
To satisfy equation, if p+q = 2, a+b should be 10.
from eq(1) if p = 1 , a = 3.
from eq.(2) if q = 1 , b = 7.
Hence p = 1 and q = 1.