Factorizing $x^5+1$ as a product of linear and quadratic polynomials.

Question

I am encountering some trouble with this question:

Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.

I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.

Can anyone help with this? Thanks.

Answer 1

Hint: Can you think of a solution to $x^5=-1$? If $a$ is a solution, $x-a$ is a linear factor, and you can use polynomial long division to factor it out.

Answer 2

Let $$s_n(x)=1+x+x^2+\dots +x^{n-1}.$$ Then $s_{n+1}(x)=x^n+s_n(x)$ and $xs_n(x)+1=s_{n+1}(x)$. Thus $$x^n+s_n(x)=xs_n(x)+1,$$ that is, $$x^n-1=(x-1)s_n(x).$$ We let $p_n(x)=s_n(-x)=1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1}.$ We then have $$(x+1)p_n(x)=1+(-1)^{n-1}x^n.$$ Set $n=5$ so that $(-1)^{5-1}=(-1)^4=1$ and $$\begin{align} x^5+1&=(x+1)p_5(x)\\ &=(x+1)(1-x+x^2-x^3+x^4) \end{align}$$

Answer 3

HINTS:

• If we want to write $z^5 + 1$ as a product of real linear & quadratic polynomials, any root of the polynomial factors must be a root of $z^5 + 1$.

• $z^5 + 1$ has four complex roots and one real root. (What are they?)

• The roots of real linear polynomials are real. The roots of real quadratic polynomials can be complex, but they must be complex conjugates of each other.

Can you use these facts to make an educated guess about what the linear and quadratic polynomials must be? If so, you can then just multiply it out and check to see if you're right.

Answer 4

$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$.

$(x+1)$ has to be a factor, because $-1$ is a root.

Now to factor the quartic, you can use the other $4$ roots of $x^5+1$. They are $-e^{\dfrac {2\pi i}5},-e^{\dfrac {4\pi i}5},-e^{\dfrac {6\pi i}5},-e^{\dfrac {8\pi i}5}$.

So $(x+e^{\dfrac {2\pi i}5})(x+e^{\dfrac {8\pi i}5})=x^2+(e^\dfrac {2\pi i}5+e^\dfrac {8\pi i}5)x+1=x^2+2\cos(\dfrac{2\pi }5)x+1$ would be one factor.

Answer 5

First you seem to know that $(x+1)|(x^5+1)$. Carry out this division with polynomial long division to get

$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$

The remaining fourth degree polynomial is symmetric or palindromic, its coefficients $1,-1,1,-1,1$ read the same forward and backwards. When this happens there will be quadratic factors that are palindromic too, so you must have

$x^4-x^3+x^2-x+1=(x^2+ax+1)(x^2+bx+1)$

Expand:

$x^4-x^3+x^2-x+1=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$

Matching like powers of $x$ gives $b=-1-a$ and $ab=-1$. Eliminating $b$ between these equations gives a quadratic equation for $a$ which you can solve.

I leave you to prove: (1) Both roots for $a$ give the same factors for $x^4-x^3+x^2-x+1$, only the order is different. (2) The quadratic factors you find have negative discriminants and thus cannot be broken down to real linear factors.

Answer 6

Here is a way of doing this without going into complex numbers.

Take the equation $x^5+1=0$ and note that if $y=\frac 1x$ then $\frac 1{y^5}+1=0$ or multiplying through by $y^5$ we have $y^5+1=0$.

If $x$ is a solution, so is $\frac 1x$. Now $x=\pm 1$ are the only solutions of $x=\frac 1x$ and $x=-1$ is a solution of the original equation. The other solutions come in reciprocal pairs whose product is $1$. This means they are roots of quadratics with constant term $1$.

We can therefore deduce a factorisation of the form $$x^5+1=(x+1)(x^2+ax+1)(x^2+bx+1)$$ where we have paired the roots as roots of the quadratics.

This can be solved in your favourite way to find $a$ and $b$.

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For example, factoring out the $x+1$ factor gives $$x^4-x^3+x^2-x+1=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1$$

Equating coefficients we obtain $a+b=-1$ and $2+ab=1$ which you should be able to solve.

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Note that these reciprocal equations can often be solved (or polynomials factored) using various techniques which reduce the degree by a factor of $2$. Here we can extract the factor $x+1$ and reduce without any issue to working with quadratics (there is no cubic to solve, as there would be with a general quartic equation).

Answer 7

If you remember the formula for the sum of a geometric progression, you will have $$\begin{align} & 1 + x + x^2 + \cdots + x^{n-1} = \frac{1-x^n}{1-x}\\ \iff & x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1) \end{align} $$ When $n = 2k+1 $ is odd, substitute $x$ by $-x$ and move the minus sign around, you will get $$x^{2k+1} + 1 = (x+1)(x^{2k} - x^{2k-1} + x^{2k-2} + \cdots - x + 1)$$ In particular, for $n = 5$, this gives you $$x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$$ This sort of factorization is something you should know. If not, you know it now.

Let's look at the coefficients of $x^k$ in second factor, they are $1, -1, 1, -1, 1$.

Did you notice it is palindromic, i.e stay the same if you reverse the order. For this sort of polynomial. If $r \ne 0$ is a root, so does $\frac1r$. When the number of this list of coefficients is odd, you can simplify the expression by grouping them in powers of $x+\frac1x$.

Let's see what happens to the factor. We have

$$\begin{align} x^4 - x^3 + x^2 - x + 1 &= x^2(x^2 - x + 1 - x^{-1} + x^{-2})\\ &= x^2((x^2 + 2 + x^{-2}) - (x + x^{-1}) - 1)\\ &= x^2((x + x^{-1})^2 - (x + x^{-1}) - 1 )\\ &= x^2\left[\left(x + x^{-1} -\frac12\right)^2 - \frac{5}{4}\right]\\ \end{align} $$ Another identity you should remember is for positive $a$, $$y^2 - a = (y + \sqrt{a})(y - \sqrt{a})$$ Apply this identity to above expression, you get

$$x^4 - x^3 + x^2 - x + 1 = x^2\left(x + x^{-1} -\frac{1+\sqrt{5}}{2}\right) \left(x + x^{-1} -\frac{1-\sqrt{5}}{2}\right)$$ Move the $x$ factor around to get rid of the $x^{-1}$ and combine with the $x+1$ factor, one obtain

$$x^5 + 1 = (x+1)\left(x^2 - \frac{1+\sqrt{5}}{2} x + 1 \right) \left(x^2 - \frac{1-\sqrt{5}}{2} x + 1 \right)$$

This is the decomposition of $x^5+1$ into product of linear and quadratic factors you are seeking.