Review question

# When is $x^2 +1$ a factor of $(3+x^4)^n-(x^2+3)^n(x^2-1)^n$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7243

## Solution

Let $n$ be a positive integer. Then $x^2 +1$ is a factor of

$(3+x^4)^n-(x^2+3)^n(x^2-1)^n$

for

1. all $n$;

2. even $n$;

3. odd $n$;

4. $n \geq 3$;

5. no values of $n$.

Note that the polynomial is a function of $x^2$.

Therefore, let the polynomial $(3+x^4)^n-(x^2+3)^n(x^2-1)^n = f(x^2)$.

From the factor theorem we know that if $x^2+1$ is a factor of $f(x^2)$, then $f(-1) = 0$.

This gives:

$(3+1)^n-(-1+3)^n(-1-1)^n = 4^n - (-1)^n4^n.$

This is zero if and only if $n$ is even, and so the factor theorem tells us the answer is (b).

If we know about complex numbers, we can say the solution to $x^2 +1 = 0$ is $x = i, -i.$

Putting either of these values into $(3+x^4)^n-(x^2+3)^n(x^2-1)^n$ for $x$ gives us $4^n - (-1)^n4^n$, as before.