Equations of Degree Higher than the Second
How to solve higher-degree equations
The solutions of third and fourth degree equations can be obtained using specific formulas. However, unlike the quadratic formula, these expressions are extremely long and impractical to memorize. Moreover, there is no general formula for solving equations of the fifth degree or higher.
To solve equations of degree higher than the second, two simplification techniques are particularly useful:
- Factorization
- Ruffini's method
Both methods reduce the degree of the original equation and often make it possible to find the solutions without resorting to more advanced formulas.
Factorization
If the equation is a polynomial of degree n of the form $$ P(x)=0 $$ then factorization consists in extracting common terms and rewriting the polynomial as a product of lower-degree factors. The solutions can then be found by applying the zero product property.
Example
Consider the cubic equation
$$ x^3 + x^2 - 2x = 0 $$
We factor out x, rewriting the expression as a product of two simpler polynomials:
$$ x \cdot (x^2+x-2) = 0 $$
We now apply the zero product property to determine the solutions.
A] The first solution is immediate.
If x = 0, then the left-hand side of the equation is zero.
$$ 0 \cdot (0^2+0-2) = 0 $$
$$ 0 \cdot (-2) = 0 $$
$$ 0 = 0 $$
B] The second and third solutions are obtained by solving the quadratic equation x2 + x - 2
$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
Given a=1, b=1, c=-2:
$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
$$ x = \frac{-1 \pm \sqrt{1^2-4(1)(-2)}}{2(1)} $$
$$ x = \frac{-1 \pm \sqrt{1+8}}{2} $$
$$ x = \frac{-1 \pm \sqrt{9}}{2} $$
$$ x = \frac{-1 \pm 3}{2} = \begin{cases} x = \frac{-1-3}{2} = -2 \\ \\ x = \frac{-1+3}{2} = 1 \end{cases} $$
The roots of the quadratic equation x2+x-2 are x = -2 and x = 1.
Both values also satisfy the original product.
Check
If x=-2 $$ x \cdot (x^2+x-2) = 0 $$ $$ (-2) \cdot ((-2)^2+(-2)-2) = 0 $$ $$ (-2) \cdot (4-2-2) = 0 $$ $$ (-2) \cdot (0) = 0 $$ $$ 0 = 0 $$ If x=1 $$ x \cdot (x^2+x-2) = 0 $$ $$ 1 \cdot (1^2+1-2) = 0 $$ $$ 1 \cdot (2-2) = 0 $$ $$ 1 \cdot (0) = 0 $$ $$ 0 = 0 $$
The solutions of the cubic equation x3 + x2 - 2x = 0 are therefore:
$$ \begin{cases} x = 0 \\ \\ x = -2 \\ \\ x = 1 \end{cases} $$
Ruffini's method
Ruffini's method provides another way to factor a polynomial of degree higher than two into a product of lower-degree factors.
Example
Consider again the cubic equation
$$ x^3 + x^2 - 2x = 0 $$
The leading coefficient (a) is equal to 1, so no preliminary manipulation is required.
One nonzero root of this polynomial is x = 1.
By Ruffini's theorem, the polynomial P(x)=x3 + x2 - 2x is divisible by (x - 1)
$$ \frac{P(x)}{(x-1)} = Q(x) $$
Thus, P(x) can be expressed as the product of (x - 1) and a quotient polynomial Q(x)
$$ P(x) = (x-1) \cdot Q(x) $$
To determine Q(x), we apply the Ruffini method
$$ \begin{array}{c|lcc|r} & 1 & 1 & -2 \\ 1 & & & \\ \hline & & & & \end{array} $$
$$ \begin{array}{c|lcc|r} & 1 & 1 & -2 \\ 1 & & 1 & 2\\ \hline & 1 & 2 & 0 & \end{array} $$
The quotient polynomial is Q(x) = x2 + 2x
So we can rewrite P(x) as
$$ P(x) = (x-1) \cdot Q(x) $$
$$ P(x) = (x-1) \cdot (x^2+2x) $$
We now apply the zero product property:
A] The first factor (x - 1) is zero when x = 1
$$ (x-1)= 0 $$
B] The second factor (x2+2x) is zero when x = 0 or x = -2
$$ x^2 +2x = 0 $$
$$ x \cdot (x+2) = 0 $$
The solutions of the original cubic equation x3 + x2 - 2x = 0 are therefore
$$ \begin{cases} x = 0 \\ \\ x = -2 \\ \\ x = 1 \end{cases} $$
These are the same solutions obtained through factorization.
Special cases
In some situations, higher-degree equations can be solved through alternative reasoning.
- Binomial equations
A binomial equation of the form $$ ax^n + b = 0 $$ can be solved by isolating the variable $$ x^n = - \frac{b}{a} $$ and then taking the nth root of both sides $$ \sqrt[n]{x^n} = \sqrt[n]{ - \frac{b}{a} } $$ $$ x = \sqrt[n]{ - \frac{b}{a} } $$Example. Consider the cubic equation $$ 2x^3 + 54 = 0 $$ Isolate the variable $$ x^3 = - \frac{54}{2}$$ $$ x^3 = - 27 $$ Take the cube root of both sides $$ \sqrt[3]{ x^3 } = \sqrt[3]{ - 27 } $$ $$ x = \sqrt[3]{ - 27 } $$ The cube root of -27 is -3 because (-3)3=-27 $$ x = -3 $$ The equation is solved.
- Trinomial equations
A trinomial equation of degree higher than the second of the form $$ ax^2n + bx^n + c = 0 $$ where the exponent 2n is double the exponent n, can be simplified by introducing the auxiliary variable z=xn $$ az^2 + bz + c = 0 $$ Once the roots in z are found, the corresponding values of x can be determined.
Example. Consider the equation $$ x^8 - 17x^4 + 16 = 0 $$ Introduce the auxiliary variable z=x4 to obtain a quadratic equation $$ z^2 - 17z + 16 = 0 $$ We now solve the quadratic in z $$ z = \frac{-(-17) \pm \sqrt{225}}{2} = \begin{cases} \frac{17 - 15}{2} = \frac{2}{2} = 1 \\ \\ \frac{ 17 + 15}{2} = \frac{32}{2} = 16 \end{cases} $$ Once the roots of z are known, we determine the values of x from z=x4 $$ x = \begin{cases} x_1^4 = 1 \\ \\ x_2^4 = 16 \end{cases} $$ $$ x = \begin{cases} x_1 = \sqrt[4]{1} \\ \\ x_2 = \sqrt[4]{16} \end{cases} $$ $$ x = \begin{cases} x_1 = \pm 1 \\ \\ x_2 = \pm 2 \end{cases} $$
And so on.
