How to factorize polynomials to the 5th degree?

Question

I have the polynomial: $$2x^5-x^4+10x^3-5x^2+8x-4$$ and I know that the final result is: $$(2x-1)(x^4+5x^2+4) = (2x-1)(x^2+1)(x^2+4)$$ But how would you do it step by step? I've seen a couple of videos and blogs about it, but they mostly use examples, where their is a common factor between the expressions but in this case their are none.

Answer 1

Every now and then, you find a polynomial of higher degree that can be factored by inspection. In this case, there's a way to just "see" one step of the factorization:

$$2x^5-x^4+10x^3-5x^2+8x-4$$

Notice that the coefficients, when grouped in pairs, are all proportional: $2, -1$ are in the same ratio as $10,-5$ and also $8,-4$. That's uncommon, but it means you can "factor by grouping":

$$(2x^5-x^4)+(10x^3-5x^2)+(8x-4)$$ $$=x^4(2x-1)+5x^2(2x-1)+4(2x-1)$$ $$=(x^4+5x^2+4)(2x-1)$$

The first step is to take the gcf out of each group, and then, since the remaining parts match ($(2x-1)$), that can be factored out. Now we're left with something that is quadratic in form, and can be factored further:

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

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Now, in most cases, nice patterns like this do not occur. Then, you're left with the rational roots theorem. From the fact that our lead coefficient is $2$ and our constant term is $4$, we conclude that any linear factor will correspond to a zero of the form $\pm\frac{p}{q}$, where $p$ is a factor of $4$ and $q$ is a factor of $2$. That gives us a few options: $\pm 4, \pm 2, \pm 1, \pm \frac12$. Trying a few of these, we see that plugging in $x=\frac12$ actually produces a value of $0$. That tells us that $(x-\frac12)$ is a factor, or clearing denominators, $(2x-1)$. You can now factor this linear piece out by long division, and end up just where we were after factoring by grouping.

Answer 2

It's a theorem that the only possible rational zeroes are $p/q$, where $p$ is a factor of $4$ and $q$ is a factor of $2$. So the only possible rational zeroes are $\pm4$, $\pm 2$, $\pm 1$ and $\pm1/2$. Try them and you see that $1/2$ is a zero. So $(2x-1)$ is a factor.

Now when you divide by $2x-1$ you get a quartic. But it's just $Q(x^2)$, where $Q$ is a quadratic. So factor $Q(x)$ as usual, replace $x$ by $x^2$ and you're done.

Answer 3

Laboring under the assumption that such a polynomial is not irreducible and can be factored (which is clear in this case since you already have a factorization), we would generally start by looking for rational roots.

The rational root theorem asserts that if $z = \frac{a}{b}$ is a rational root of the polynomial (in lowest terms), then $b$ must divide the leading coefficient, and $a$ must divide the constant term. This gives us several potential rational roots: $$ \left\{ \pm 1, \pm 2, \pm 4, \pm \frac{1}{2} \right\}. $$ By trial and error, it is possible to determine that the only one of these that is, in fact, a root is $\frac{1}{2}$. From this, we know that $$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right) q(x), $$ where $q$ is a polynomial of degree 4. Via polynomial long division, Horner's algorithm / synthetic division, or some other technique, we can deduce that $q(x) = 2x^4 + 10x^2 + 8$, thus $$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right)\left( 2x^4 + 10x^2 + 8 \right). $$ The last term is quadratic in $x^2$. Via the quadratic formula, completing the square, or some other technique, we can determine that it, too, factors. For example, we can "factor by grouping:" $$ 2x^4 + 10x^2 + 8 = 2x^4 + 8x^2 + 2x^2 + 8 = 2x^2(x^2 + 4) + 2(x^2+4) = (2x^2 + 2)(x^2 + 4). $$ Note that both of these factors are irreducible over $\mathbb{R}$ (they have complex roots, but no purely real roots), hence we can stop here. Putting all of the pieces back together, we get $$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right) (2x^2 + 2)(x^2 + 4), $$ which is equivalent to the factorization you gave.