If on dividing the polynomial #x^4-x^3-13x^2+sx+t# by #(x+3)(x+4)# remainder is #0#, find the value of #s# and #t#?
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#frac{x^4 - x^3 - 13x^2 + sx + t}{x^2 + 7x + 12} = x^2 + frac{- 8x^3 - 25x^2 + sx + t}{x^2 + 7x + 12}#
#= x^2 - 8x + frac{(-25+56)x^2 + (96 + s)x + t}{x^2 + 7x + 12}#
#= x^2 - 8x + (31) + frac{(-31*7+96 + s)x + (-31*12 + t)}{x^2 + 7x + 12}#
According to Remainder theorem if #(x-a)# is a factor of the polynomial #f(x)#, then #f(a)=0#.
Now as dividing the polynomial #x^4-x^3-13x^2+sx+t# by #(x+3)(x+4)# results in #0# as remainder, both #(x+3)# and #(x+4)# are a factor of the polynomial #f(x)=x^4-x^3-13x^2+sx+t#
and as per remainder theorem #f(-3)=0# and #f(-4)=0# i.e.
#(-3)^4-(-3)^3-13(-3)^2+s(-3)+t=0#
or #81+27-117-3s+t=0#
i.e. #-3s+t=9# ...................(1)
and #(-4)^4-(-4)^3-13(-4)^2+s(-4)+t=0#
or #256+64-208-4s+t=0#
i.e. #-4s+t=-112# ...................(2)
Subtracting (2) from (1), we get
#-3s+t-(-4s+t)=9-(-112)#
or #-3s+t+4s-t=9+112# or #s=121#
and putting this in (1), we get
#-3xx121+t=9# or #-363+t=9# i.e. #t=372#
