Solving quartic equations

Question

Given the following quartic equation:

$$x^4-2x^3-7x^2+8x+12=0$$

Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?

Answer 1

Hint: $x = 2$ is a root. Can you take it from here? in general, there are some tricks out there, for example if you have the type I: $x^4+ ax^3+ bx^2+ ax + 1 = 0$, then you divide both sides by the middle $x^2$ and put $y = x+\dfrac{1}{x}$, reducing it to a quadratic equation immediately.

Answer 2

Notice, $$x^4-2x^3-7x^2+8x+12=0$$ substituting $x=-1$ we get $$(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12=0\iff 0=0$$ Hence, $x=-1$ is a root of given quartic equation i.e. $(x+1)$ is a factor of $x^4-2x^3-7x^2+8x+12$, now we have $$(x+1)(x^3-3x^2-4x+12)=0$$
further factorizing $x^3-3x^2-4x+12=(x-3)(x^2-4)$ $$(x+1)(x-3)(x^2-4)=0$$ $$(x+1)(x-3)(x-2)(x+2)=0$$

Hence, the roots are $$x=\color{red}{-2, -1, 2, 3}$$

Answer 3

As an A level student, I would first look at this with the Remainder Theorem, which states that if we sub in a value, say 1, and we get zero, then (x-1) is a factor.

Let $$p(x)=x^4-2x^3-7x^2+8x+12$$ Sub in -1: $$p(-1)=(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12 = 0$$ Therefore, (x + 1) must be a factor. We can then use algebraic division to gain a cubic equation. After doing the division, we are left with: $$p(x)=(x+1)(x^3-3x^2-4x+12)$$ We can then use the remainder theorem to try and factorise the cubic. Subbing in -2, we get another 0. Therefore, (x+2) is a factor of the cubic. Long division gives that p(x) equals: $$p(x)=(x+1)((x+2)(x^2-5x+6))$$ The quadratic can be factorised to give: $$p(x)=(x+1)(x+2)(x-3)(x-2)$$ Since this equals 0, then: $$x+1 = 0 \implies x = -1$$ $$x+2 = 0 \implies x = -2$$ $$x-3 = 0 \implies x = 3$$ $$x-2 = 0 \implies x = 2$$ To confirm, a graph gives:

Answer 4

Are you familiar with the rational roots theorem? It says that any rational roots of a polynomial with integer coefficients must have the form $\frac{p}{q}$ where $p$ is a factor of the polynomial's constant term and $q$ is a factor of its leading term.

The constant term of your polynomial is $12$ and its leading term is $1$, so the possibilities for rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$, and $\pm 12$. Note that these are only possibilities. You have to test them to see if they work. If you test all the possibilities, you'll find that there are four which work ($-1, 2, -2$, and $3$). Since any quartic polynomial has at most four real roots, you can be sure that you've found all of them.

Unfortunately, many quartic polynomials have roots which are not rational. For example, suppose that you can only find two roots $a$ and $b$ using the rational roots theorem. You could then divide the polynomial by the factor $x - a$ and then again by the factor $x-b$ to produce a quadratic equation (you can do this because of the remainder theorem, mentioned in another answer). Then use the quadratic formula to find the remaining roots.

So, basically, this technique is only useful when your polynomial has at least two rational roots.

There are techniques for solving any quartic equation, but they are quite advanced. You don't see them until you study field theory in depth. You might encounter the topic as an advanced undergraduate, but even then I think it's usually skipped.

Answer 5

For reference and examples on Quartic equations check out this site: The Quartic Formula

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To solve a quartic equation Ludovico Ferrari used a very similar method as for the cubic equation.(The Cubic Formula)

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I. The first objective is to find two roots of equation (1). Once we have them the other two roots (real or complex) can be found by polynomial division and the quadratic formula.

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II. To find the two roots we use a substitution:

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to the quartic equation (1) to obtain:

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Multiplying out and simplifying, we obtain the "depressed" quartic (no cubed term)

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For example:

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III. Solving the depressed quartic.

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We will first move the y-term to the other side:

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Next, we "complete the square" on the left side, but we want to end up with

To accomplish that, we add the term:

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on both sides:

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Simplifying the left side results in

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Now we use a new unknown z, we want to change the left side of (5) to

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which means to add :

To both sides of (5) to obtain:

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The left side is a "perfect square". The next step will be to determine z so that the right side looks like a perfect square as well. For that one needs to compute the determinant of the RHS. If the determinant is 0 we have a repeated root, which means a we have a perfect square. With this we get a value for z and the rest of the solution is plugging it back to our final equation and solve the quadratic. The other two roots can be found by polynomial long division, and application of the quadratic formula.