# 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 $$f(x)$$.

Note by Ayush G Rai
2 years, 1 month ago

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

Staff - 2 years, 1 month ago

- 2 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 $$\boxed {-x^5}.$$

- 2 years, 1 month 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.

Staff - 2 years, 1 month ago

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

- 2 years, 1 month ago

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.

Staff - 2 years, 1 month ago

What have you tried? What are your thoughts​?

Staff - 2 years, 1 month ago

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

- 2 years, 1 month ago