What is the value of b?

#P(x)=x^4-x^3-3bx^2+12b^4#
#D(x)=(x+3b)#
#R(x)=93#

1 Answer
Oct 31, 2017

#b = {(-1 or 1, "Reals"),(i or -i, "Complex"):}#

Explanation:

According to the Remainder Theorem, the remainder when dividing a polynomial #P(x)# by a divisor #D(x)# is equal to the value of the root found when you set the divisor equal to zero.

In the context of this problem, this means we set the divisor #D(x) = 0#, solve for the root #x_0#, and then we know that the remainder #R(x) = P(x_0)#.

#D(x) = 0#

#x + 3b = 0 => x = -3b#

Since the root is #-3b#, we know that #P(-3b) = R(x)#, or #P(-3b) = 93#:

#P(-3b) = 93#

#(-3b)^4 - (-3b)^3 - 3b(-3b)^2 + 12b^4 = 93#

#81b^4 + 27b^3 - 27b^3 + 12b^4 = 93#

#93b^4 = 93#

#b^4 = 1#

#:. b = -1 or 1#

Test -1

#P(x) = x^4 - x^3 + 3x^2 + 12#

#D(x) = x - 3#

Using synthetic division to test for the remainder:

#3__| color(white)("aaaaa")1color(white)("aaaaa")-1color(white)("aaaaa")3color(white)("aaaaa")0color(white)("aaaaa")12#
#color(white)("aaaaa")underline(color(white)("aaaaaaaaaa")3color(white)("aaaaa")6color(white)("aaaaa")27color(white)("aaaaa")81)#
#color(white)("aaaaaaa")1color(white)("aaaaaaa")2color(white)("aaaaa")9color(white)("aaaaa")27color(white)("aaaaa")color(green)(93)#

Test 1

#P(x) = x^4 - x^3 - 3x^2 + 12#

#D(x) = x + 3#

Using synthetic division to test for the remainder:

#-3__| color(white)("aaaaa")1color(white)("aaaa")-1color(white)("aaaaa")-3color(white)("aaaaa")0color(white)("aaaaa")12#
#color(white)("aaaaaaaa")underline(color(white)("aaaaa|a")-3color(white)("aaaaaa")12color(white)("aaa")-27color(white)("aaaa")81)#
#color(white)("aaaaaaaaa")1color(white)("aaaa")-4color(white)("aaaaaaa")9color(white)("aaa")-27color(white)("aaaa")color(green)(93)#

Final Note

There are complex solutions possible as well, which would result in a polynomial with complex coefficients. I omitted those in this solution as I assumed the problem was seeking a real polynomial. (Add #b = +-i# for the other solutions)