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)