When the polynomial f(x)=px^2+qx+r, where p,q and r are condinates is divide by (x+3) and (x-2), the remainder are -12 in each case.if x+l is a factor of f(x), find f(x) find the zeros of f(x)
when f(x)=px^2+qx+r is divided by x+3, the remainder is -12
---> f(-3) = 9p - 3q + r = -12
when f(x)=px^2+qx+r is divided by x-2, the remainder is -12
---> f(2) = 4p + 2q + r = -12
subtract these two:
5p - 5q + 0 = 0 , p - q = 0
p = q
also x+1 is a factor, so
f(-1) = p - q + r = 0
but p-q = 0 , so r = 0
back to 4p + 2q + r = -12
4p + 2p + 0 = -12
6p = -12
p = -2
f(x) = -2x^2 - 2x
check:
f(1) = 0-0 = 0 , so x-1 is a factor, that was given
f(2) = -2(4) - 4 = -12, good!
f(-3) = -2(9) + 6 = -12, all is good!