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#?

2 Answers
Vinícius Ferraz · Shwetank Mauria
Jul 5, 2017

#s = 121, t = 372#

Explanation:

#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}#

Shwetank Mauria
Jul 5, 2017

#s=121# and #t=372#

Explanation:

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#

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