Fifth Degree Polynomial!

Let $f(x)$ be a $5^\text{th}$ degree polynomial such that $f(x) + 1$ is divisible by $(x-1)^3$ , and $f(x)-1$ is divisible by $(x+1)^3$ . Find $f(x)$ .

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

We have $f(x) + 1 = (x-1)^3 g(x)$ and $f(x) -1 = (x+1)^3h(x)$ . Thus, we want to find degree 2 polynomials such that $g(x) (x-1)^3 - h(x) (x+1)^3 =2$ .

This is just bezout's identity applied to polynomials.

$(x+1)^3 = 1 \times (x-1)^3 + (6x^2 + 6)$
$(x-1)^3 = (\frac{1}{6}x-\frac{1}{2}) \times (6x^2 + 6) +(2x+2)$
$(6x^2 +6) = (3x-3)(2x+2) + 12$

So we apply the backwards step of euclidean algorithm to obtain

$2 = 1 \times (x^2 + 1) - (\frac{x-1}{2}) \times (2x+2) \\ = 1 \times (x^2 + 1) - ( \frac{x-1}{2} ) \times [ (x-1)^3 - (x-3)(x^2+1) ] \\ = ( \frac{ x^2 -4x+5} { 2} ) \times (x^2 + 1) - ( \frac{x-1}{2} ) \times (x-1)^3 \\ = ( \frac{ x^2 -4x+5} { 2} ) \times \frac{1}{6} [ (x+1)^3 - (x-1)^3] - ( \frac{x-1}{2} ) \times (x-1)^3 \\ = \frac{ x^2-4x+5}{12} \times (x+1)^3 - \frac{x^2+2x-1}{12} \times (x-1)^3 \\$

Thus, $f(x) = \frac{ x^2-4x+5}{12} \times (x+1)^3 -1$ .

Calvin Lin Staff - 5 years ago

Totallly forgot about euclidian's algorithm.

Ian Limarta - 4 years, 11 months ago

What have you tried? What are your thoughts?

Calvin Lin Staff - 5 years ago

I'm trying it out.I have no idea right now.

Ayush G Rai - 5 years ago

We know that $(x-1)$ is a factor of $f(x)+1.$ Using factor and remainder theorem,we get
$f(1)+1=0$
$f(1)=-1$
We know that $(x+1)$ is a factor of $f(x)-1.$ Using factor and remainder theorem,we get
$f(-1)-1=0$
$f(-1)=1$
while seeing the two equations we can guess that the leading co-efficient is 1 and that the polynomial is $\boxed {-x^5}.$

Ayush G Rai - 5 years ago

Unfortunately $-x^5 + 1$ is not a multiple of $(x-1)^3$ .

You had a good start, applying remainder factor theorem. You applied it to just $(x-1)$ and so you got 1 condition. You should also apply it to $(x-1)^2$ and $(x-1)^3$ to get more conditions. Ditto for $(x+1)^3$ . This gives you 6 conditions, and we have 6 unknowns, so we should be able to solve that.

Calvin Lin Staff - 5 years ago

But my answer is $-x^5$ and not $-x^5+1.$

Ayush G Rai - 5 years ago

@Ayush G Rai – If your answer is $f(x) = - x^5 + 1$ , the question states that " $f(x) + 1$ is divisible by $(x-1)^3$ ", which is not true.

Calvin Lin Staff - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$