If k is a zero of the polynomial function

Question

If k is a zero of the polynomial function
f(x)= ax⁴ – bx³ + cx² – bx + a, then …… is also a zero of f

1) k + 1
2) k – 1

3) 1/ k
4) –k

Answer 1

-k is NOT a solution. By counterexample:

Let k=1, and a=1, b=1, c=0

Then f(x) = f(k) = f(1) = 1(1)⁴ - 1(1)³ + 0(1)² - 1(1) + 1 = 1 - 1 + 0 - 1 + 1 = 0

So, k [=1] is a zero of the function (when a=1, b=1, c=0).

Now, -k = -1, and a=1, b=1, c=0 (same!)

Then f(x) = f(-k) = f(-1) = 1(-1)⁴ - 1(-1)³ + 0(-1)² - 1(-1) + 1 = 1 + 1 + 0 + 1 + 1 = 4

So, -k [=-1] is NOT a zero of the function (when a=1, b=1, c=0 all the same)

1/k is a Solution. Proof:

For ANY a, b, and c, (a≠0, as the leading coefficient of the polynomial),

Given: For ANY k being a zero of the function: f(x) = f(k) = ak⁴ - bk³ + ck² - bk + a = 0

Then f(x) = f(1/k) = a(1/k)⁴ - b(1/k)³ + c(1/k)² - b(1/k) + a =

(a(1/k)⁴ - b(1/k)³ + c(1/k)² - b(1/k) + a) * k⁴

(k≠0) = -------------------------------------------------------- =

k⁴

a - bk + ck² - bk³ + ak⁴ 0 (same polynomial with reverse order of terms)

= ------------------------------- = ------ = 0

k⁴ k⁴

In this multiple choice, the answer is 3) 1/ k.

If complete statement was required:

Answer: If k is a zero of the function, then 1/k is also a zero of the function, for any k ≠ 0.

Answer 2

ak^4 - bk^3 + ck^2 - bk + a = 0

 

k^4 - (b/a)k^3 + (c/a)k^2 - (b/a)k + 1 = 0

 

k^4 + (b/a)(-k^3) + (c/a)k^2 + (b/a)(-k) + 1 = 0

 

(-k)^4 + (b/a)(-k)^3 + (c/a)(-k)^2 + (b/a)(-k) + 1 = 0

 

So -k is also a solution