How can I find the roots of the quartic polynomial $2x^4 −3x^3 +5x^2 +6x−4$?

Question

I know that for a quartic polynomial $p(x)=ax^4+bx^3+cx^2+dx+e$ with $a=1$ one of the roots is a factor of $p(x)$. However here $a\neq1$, so I presume there is a trick to simplify this polynomial?

Answer 1

A general statement for every polynomial $p(x)$ is: If you find a root $\alpha$ such that $p(\alpha)=0$, then you can rewrite your polynomial as $p(x)=(x-\alpha)\cdot \tilde{p}(x)$ where $\tilde{p}$ is obtained by dividing $p$ by $(x-\alpha)$.

So we have to find the roots of your function. If $f(x)=2x^4-3x^3+5x^2+6x-5$ is your function, there exist a formula to find exact roots but it is quite complicated and rarely what is asked. Here I would just test some easy values of $x$ and see how it turns out: $$f(0)=-4\qquad f(1)=2-3+5+6-4=6$$ And with some luck: $$f(-1)=2(-1)^4-3(-1)^3+5(-1)^2+6(-1)-4=2+3+5-6-4=0$$ So we have our first root $\alpha_1=-1$

Then we divide $f(x)$ by $(x-\alpha_1)=(x+1)$ and we get the equality: $$f(x)=(x+1)(2x^3-5x^2+10x-4)$$ And you could try to simplify it further by finding other roots for the second polynomial (there exist one more root on the real axis) but as I don't know how far you need to go I'll leave that to you.

Answer 2

As others posted, sometimes just trying some obvious numbers gets you a lucky answer. Someone mentioned the rational root theorem, which says that a rational root $p/q$ must have the coefficient on the highest exponent divisible by p, and the coefficient on the lowest (i.e., the constant) must be divisible by q. With 2 and -4, your set of possible rational roots is:

$$ \left\{ 0, \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{4} \right\} $$

So take your lucky guesses among those :)

Answer 3

Use the Rational Root Theorem.

We have to test the following eight values: $$1, -1, 2, -2, 4, -4, \frac{1}{2}, -\frac{1}{2}$$

It seems like a lot to test, and frankly it is, but luckily $-1$ works. That means that a factor of the polynomial is $x+1$. (Why? Because then the polynomial is $$(x+1)\cdot(\text{some other stuff}) = 2x^4-3x^3+5x^2+6x-4$$

and we know that $x = -1$ is a root, so then the LHS equals $0$. So does the RHS. Therefore, we divide the RHS by $x+1$ to find the $\text{some other stuff}$, which will be a cubic polynomial.)

We divide $2x^4-3x^3+5x^2+6x-4$ by $x+1$. (You can do this using synthetic division, or standard polynomial division.) You should get $2x^3-5x^2+10x-4$.

We need to find one more factor, because then we can use the quadratic formula to finish it off. Note that any negative value would not work, since then $2x^3$ would be negative, and same with $-5x^2$, $10x$, and $-4$.

So we just need to test $\frac{1}{2}, 1, 2,$ and $4$. We can find that $x = \frac{1}{2}$ works. (If you calculated $x = 1$ and $x = 2$ first, you can see that as $x$ increases, so does $2x^3-5x^2+10x-4$, above $0$. So instead of $4$, you would test $\frac{1}{2}$ next.)

Therefore, $2x-1$ is a factor, and dividing it from $2x^3-5x^2+10x-4$ should get you $x^2-2x+4$.

From here, just use the quadratic formula. You should get $x = 1 \pm i\sqrt{3}$.

Therefore, the four roots are $$\frac{1}{2}, -1, 1+i\sqrt{3}, 1-i\sqrt{3}$$

Answer 4

Well there is a formula to solve order $4$ equations (and cubic equations and quadratics) but it's quite tricky. My approach is to substitute in values $v$ for $x$ until you find one that makes the whole expression equal to zero. In this case $(x-v)$ will be a factor.

If we set $x=-1$ we see that the expression is equal to $2+3+5-6-4$ which is $0$ and so $(x+1)$ is a factor. Now let's divide the expression by $(x+1)$.

I don't know how to show this in Latex but the answer is $2x^{3}-5x^{2}+10x-4$. When $x$ is set to $0$ then this cubic is equal to $-4$ and when $x$ is set to $1$ it is $3$. This tells us that it crosses the $x$-axis between $0$ and $1$. And approximating the curve as a straight line suggests the value $x=0.5$ as a solution of the equation. Let's try this and we get $0.25-1.25+5-4=0$ . Yeah, it works! and so $x=1/2$ is a solution and hence $2x-1$ is a factor. Again let's divide the cubic by $2x-1$ and we get $x^{2}-2x+4$. Now does this factorise? Well not in the real numbers because the discriminant $b^{2}-4ac$ (from the quadratic formula $x=(-b \pm \sqrt {b^{2}-4ac}) /2a$) is equal to $-12$ and is negative hence we can't take its square root and so there are no more real number solultions.

If you want them, the complex number solutions will be $1+\sqrt3i$ and $1-\sqrt{3}i$. This will then give two factors of the quadratic (and hence the initial equation), they are $x-1-\sqrt{3}i$ and $x-1+\sqrt{3}i$.

Answer 5

A polynomial where the coefficient of the highest power is $1$ is called monic. However, given a non-monic polynomial, we can just divide by the leading coefficient to get a different polynomial with the same roots. For example, we can divide $p(x)=3x^2+6x+9$ by $p(x)/3 = x^2 + 2x + 3$. It may be helpful to think of this as factoring out a constant. So in the previous example, we could also say $p(x)=3(x^2+2x+3)$.

With polynomials, just like with numbers, you can do division with remainder, and if you divide a polynomial $p(x)$ by a polynomial $q(x)$ to get a quotient $a(x)$ and a remainder $b(x)$, that means that $p(x)=a(x)q(x)+b(x)$ (where the degree of $b$ is smaller than the degree of $q$). We don't need anything to be monic here, although if $p$ and $q$ have whole number coefficients and $q$ is monic, then $a(x)$ and $b(x)$ will have whole number coefficients too, but if $q(x)$ is not monic, then $a(x)$ and $b(x)$ might only have rational coefficients.

If we divide $p(x)$ by the polynomial $q(x)=(x-\alpha)$, which is of degree $1$, then the remainder will have degree less than $1$, which means the remainder is actually a constant. But what constant is it? If $p(x)=a(x)(x-\alpha)+c$, then we can plug in $x=\alpha$ to get $p(\alpha)=a(\alpha)(\alpha-\alpha)+c=a(\alpha)(0)+c=0+c=c$. This tells us that we have remainder $0$ if and only if $p(\alpha)=0$. So linear factors of $p(x)$ are "the same as" roots of $p(x)$. Note that this doesn't require $p(x)$ to be monic. It also puts no requirements on the degree of $p(x)$.

Fining roots or factors of a quartic equation is hard. You don't have to have any real roots at all, and while you might factor as a product of two quadratic polynomials, you also might not. There is a formula like the quadratic formula for 4th degree equations, but it is literally pages long. Instead, there are three common ways to try to find roots of quartic equations: try to factor by grouping, look for your quartic to fit into a pattern you have seen, e.g., a perfect $4$th power, a quadratic in $x^2$, or apply the rational roots theorem to find a rational root, which then gives you a linear factor you can divide by. However, the rational roots theorem only tells you that if there is a rational root then it is in a small list of numbers. It can be viewed as a fancy form of guessing and testing.

By the fundamental theorem of algebra, you know that you will have 4 roots to every quartic (counted with multiplicities), but they might all be complex, even for a very nice looking quartic. Complex roots can be difficult to think about. Depending on what you are doing, the fact that these roots exist might not be useful to you, especially if you have no good way to find them.

There are more advanced things that you can do to try to factor quartics, such as using Eisenstein's critereon to show that your quartic doesn't even factor into two quadratics with integer coefficients. There is a lot of lovely math associated with the problem of factoring, which is a really hard problem in general. However, you shouldn't worry about the more advanced techniques for now.