Algebra Question

Question

Find the sum of all k for which x⁵ + kx⁴ − 6x³ − 15x² − 8k³x − 12k + 21 leaves a remainder of 23 when divided by x + k.

Answer 1

By remainder theorem remainder r = f(-k) = - k⁵ + k⁵ + 6k³ - 15k² + 8k⁴ - 12k + 21 = 23;

8k⁴ + 6k³ - 15k² - 12k - 2 = 0;

By Vietas formula k₁ + k₂ + k₃ + k₄ = - 6/8 = - 3/4

Answer 2

First write the polynomial division as a quotient-and-remainder. For some unknown (and irrelevant) quotient polynomial q(x), we have

x⁵+kx⁴-6x³-15x²-8k³x-12k+21=q(x)(x+k)+23

This equation holds for all x, so we can plug in x=-k to make the q-term vanish, and we get

(-k)⁵+k(-k)⁴-6(-k)³-15(-k)²-8k³(-k)-12(-k)+21=23

Once we (carefully) simplify everything and collect all our like terms, we end up with

8k⁴+6k³-15k²+12k-2=0

Now fear not; we don't have to solve a 4th-degree polynomial. Your question asks for the sum of all k's which satisfy this equation, and this sum is simply -3/4.

If you have a factored polynomial (x-a)(x-b)(x-c)(x-d)=0 with a leading coefficient of 1, the solutions are of course a, b, c, d. Now expand out this polynomial, and look at the fully-simplified x³ term: it's -(a+b+c+d), the negative sum of the solutions.

So the sum of the roots of our k-polynomial will be the k³ term, if the leading coefficient is 1. But that's an easy fix; divide by 8 to make the leading coefficient 1, and this turn the k³ coefficient into 6/8, or 3/4. The desired sum is the negative of this number, -3/4.