 # Quartic Equations

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Quartic equations in one variable are equations in which the variable is raised to the 4th power in at least one term. In general, they are equations of the form ax^4 + bx^3 + cx^2 + dx + e = 0, where a ≠ 0. Quartic equations are sometimes also referred to as bi-quadratic equations (but I will refer to them as quartic equations since I consider biquadratic equations to be a special form of quartic equation as we shall see a little later).

As mentioned in the previous lesson on cubic equations, the factorization method can be used to solve quartic equations just like it can be used to solve quadratic or cubic equations. Instead of applying it once (as in quadratic equations) or twice (as in the case of cubic equations), we need to apply the method thrice in the case of quartic equations to solve them. In general, the number of applications of the factorization method is one less than the degree of the equation that we are trying to solve.

Since the factorization method is familiar to us already, this lesson will cover the basics needed to set the stage for their application to quartic equations. We will then jump into some applications of the method. I am not going to spend too much time on quartic equations because we don’t encounter many of them commonly. That is also the reason I am not going to extend the factorization method to higher order equations with separate lessons though it is equally applicable to such equations.

In general, suppose we are given a quartic equation as below:

ax^4 + bx^3 + cx^2 + dx + e = 0

Then, our aim in the factorization exercise is to convert it into an equation of the form below:

ax^4 + fx^3 + gx^3 + hx^2 + jx^2 + kx + mx + e = 0

The coefficients of the new equation have to satisfy the following conditions:

a/f = g/h = j/k = m/e
f + g = b
h + j = c
k + m = d

Once again, we get a system of 6 non-linear equations in 6 unknowns which we can solve for the coefficients directly, given enough time. But it is more convenient (in most cases where we know the equation has real, rational roots) to solve the system using trial and error.

One application of the factorization method will usually produce one root of the equation, and a cubic equation. We can then solve the cubic equation by applying the factorization method once again to it. The procedure is the same as the application of the method to quartic equations and was explained in full in the previous lesson. This will then produce another root of the equation, and a quadratic equation, which we can choose to solve by any method that is convenient to us (factorization, quadratic formula, etc.).

Before we go any farther, notice the following properties of quartic equations:

• If the quartic equation lacks a constant term (that is, it is of the form ax^4 + bx^3 + cx^2 + dx = 0), then x = 0 is the first root of the equation. We are then left with the cubic equation ax^3 + bx^2 + cx + d = 0 to solve for the other three roots
• If a + b + c + d + e = 0 in a quartic equation, x = 1 is the first root of the equation
• Similarly, if a + c + e = b + d, then x = -1 is the first root of the equation
• If both b and d are zero, then the equation is of the form ax^4 + cx^2 + e = 0 (it is more appropriate to refer to this form of quartic equation as a biquadratic equation rather than one with all the coefficients being non-zero). We can then substitute y = x^2 to get the quadratic equation ay^2 + cy + e = 0. After finding the roots of the quadratic equation, take their square roots to find the roots of the original quartic equation

We will now solve a few quartic equations to get a general feel for the factorization technique as well as the application of the observations above.

Let us start with the equation 2x^4 + 5x^3 – 11x^2 – 20x + 12 = 0. We see that a = 2, b = 5, c = -11, d = -20 and e = 12. In preparation for factorization, we then write the following equations:

2/f = g/h = j/k = m/12
f + g = 5
h + j = -11
k + m = -20

As in the case of cubic equations, I find it easier to fix the first ratio (in this case, 2/f) first. That fixes f, which in turn fixes g (because of the first sum), then h (because of the second ratio), then j (because of the second sum), then k (because of the third ratio) and lastly m (because of the third sum). If the last ratio (in this case, m/12) turns out to be the same ratio as the first, 2/f, then the factorization is done. Otherwise, I try a different value for the first ratio, 2/f.

Trials with some values of 2/f are listed below (the value of the last ratio is in parentheses at the end of each line):

1: 2/2 = 3/3 = -14/-14 ≠ -6/12 (-0.5)
-1: 2/-2 = 7/-7 = -7/7 ≠ -24/12 (-2.0)
0.5: 2/4 = 1/2 = -13/-26 = 6/12 (0.5)

Thus, we have found the correct factorization to proceed forward:

2x^4 + 4x^3 + x^3 + 2x^2 – 13x^2 – 26x + 6x + 12 = 0 implies
(2x^3)(x + 2) + (x^2)(x + 2) – (13x)(x + 2) + 6(x + 2) = 0 implies
(2x^3 + x^2 – 13x + 6)(x + 2) = 0

Thus, the first root of the equation is x = -2. Notice that when -2 is a root of the given equation, a ratio of 0.5 works. Thus, if one of the roots of the equation is r, then a ratio of -1/r will work in the factorization.

We are now left with the cubic equation 2x^3 + x^2 – 13x + 6 = 0. We see that a = 2, b = 1, c = -13 and d = 6. We set out the following equations in preparation for the factorization:

2/e = f/g = h/6
e + f = 1
g + h = -13

A few trial and error factorizations are shown below (as always, I have fixed the first ratio first, and carried it through the ratios and sums all the way to h. The ratio h/6 is then reported in parentheses at the end of each line):

1: 2/2 = -1/-1 ≠ -12/6 (-2)
-1: 2/-2 = 3/-3 ≠ -10/6 (-1.6667)
0.5: 2/4 = -3/-6 ≠ -7/6 (-1.1667)
-0.5: 2/-4 = 5/-10 = -3/6 (-0.5)

Thus, we get the alternate, factorizable form of the equation as below:

2x^3 – 4x^2 + 5x^2 -10x – 3x + 6 = 0 implies
(2x^2)(x – 2) + (5x)(x – 2) – (3)(x – 2) = 0 implies
(2x^2 + 5x – 3)(x – 2) = 0

This gives us the second root of the equation as x = 2. Here, once again, we find that for a root of 2, a ratio of -0.5 (which is -1/root) works. It also leaves us with the quadratic equation 2x^2 + 5x – 3 = 0 to solve. We see easily that the factorizable form of the quadratic equation is:

2x^2 – x + 6x – 3 = 0

This then gives us the two remaining roots of the equation, x = -3 and x = 0.5. Thus, the four roots of our equation are x = 0.5, x = -3, x = 2 and x = -2.

To perform a sanity check on our factorization, let us apply the gunitasamuccaya samuccayagunita sutra. The factors we found are (x + 2), (x – 2), (x + 3), and (2x – 1). The product of the sums of the coefficients is 3*-1*4*1 = -12. The sum of the coefficients in the product is 2 + 5 – 11 – 20 + 12 = -12. Thus, we can be reasonably confident in our factorization.

Now, consider the equation 2x^4 – 3x^3 – 7x^2 + 12x – 4 = 0. We notice immediately that 2 – 3 – 7 + 12 – 4 = 0. Thus, we can conclude that x = 1 is one of the roots of this equation. We then use the knowledge that if r is a root of the equation, then -1/r will work as a valid ratio for factorizing the equation. This knowledge enables us to rewrite the given equation as below:

2x^4 – 2x^3 – x ^3 + x^2 -8x^2 + 8x + 4x – 4 = 0 implies
(2x^3)(x – 1) – (x^2)(x – 1) – (8x)(x – 1) + 4(x – 1) = 0

We then extract the first root, x = 1, and are left with the equation 2x^3 – x^2 – 8x + 4 = 0 as the residual cubic equation to solve. By simple inspection of the above equation, we can see that we don’t need to actually factorize this equation using our normal method. The equation as written above already satisfies the madhyamadhyena adhyamantyena sutra (the product of the two middle terms is equal to the product of the first and last terms). Instead, we can simply rewrite this equation as below:

(x^2)(2x – 1) -4(2x – 1) = 0 implies
(x^2 – 4)(2x – 1) = 0

This gives us the second root of the equation, x = 1/2. We are left with a residual quadratic equation, x^2 – 4 = 0. We can readily see that the two roots of this equation are x = 2 and x = -2. Thus, the four roots of the given equation, 2x^4 – 3x^3 – 7x^2 + 12x – 4 = 0, are x = 1, x = 0.5, x = 2 and x = -2.

Now, consider the equation 2x^4 – 7x^3 – 2x^2 + 13x + 6 = 0. We notice that 2 – 2 + 6 = -7 + 13. Thus, from one of our observations, we can deduce that x = -1 is one of the solutions of the equation. Thus our first root is x = -1. Because of this, we also deduce that 1 is a valid ratio for rewriting the given equation in factorizable form (1 = -1/-1). Thus, we rewrite the equation as below:

2x^4 + 2x^3 – 9x^3 – 9x^2 + 7x^2 + 7x + 6x + 6 = 0 implies
(2x^3)(x + 1) – (9x^2)(x + 1) + (7x)(x + 1) + (6)(x + 1) = 0

This factorizes out our first root, and leaves us with the residual cubic equation 2x^3 – 9x^2 + 7x + 6 = 0. We then try to factorize this equation by trial and error as below:

1: 2/2 = -11/-11 ≠ 18/6
-1: 2/-2 = -7/7 ≠ 0/6
0.5: 2/4 = -13/-26 ≠ 33/6
-0.5: 2/-4 = -5/10 = -3/6

Thus, our cubic equation can be rewritten as below:

2x^3 – 4x^2 – 5x^2 + 10x – 3x + 6 = 0 implies
(2x^2)(x – 2) – (5x)(x – 2) – (3)(x – 2) = 0 implies
(2x^2 -5x -3)(x – 2) = 0

This gives us the second root of our equation, x = 2, and the residual quadratic equation, 2x^2 – 5x – 3 = 0. We can then rewrite this equation in factorizable form as below:

2x^2 + x – 6x – 3 = 0 implies
x(2x + 1) -3(2x + 1) = 0 implies
(x – 3)(2x + 1) = 0

This then gives us the remaining two roots of the equation, x = 3 and x = -0.5.

For an example of the application of quadratic factorization to a biquadratic equation, you can read the full lesson on my blog here.

Hope this, and the previous lesson, combined have given you a good introduction to the use of factorization to solve equations of practically any degree. In this lesson, we also saw how to use the factorization method to divide a given equation by a given root to get the residual equation. Hope you will take the time to practice some of these methods so that you become completely proficient at these techniques. Good luck, and happy computing!