Factorise a Cubic Easily

The process of factorising a cubic equation can seem daunting at first, but it can be made easier by following a few simple steps and understanding the underlying principles. Cubic equations are polynomial equations of degree three, which means the highest power of the variable (usually x) is three. They have the general form of (ax^3 + bx^2 + cx + d = 0), where (a), (b), (c), and (d) are constants, and (a \neq 0).

Understand the Basics

Before diving into factorising cubic equations, it’s essential to understand the basics of polynomial factorisation. Factorisation involves expressing a polynomial as a product of simpler polynomials, called factors. For cubic equations, we aim to break them down into the product of a linear factor (if possible) and a quadratic factor, which can then be further factorised using quadratic formulae or other techniques if necessary.

Step 1: Identify Possible Rational Roots

The first step in factorising a cubic equation is to try and find a rational root. According to the Rational Root Theorem, any rational root, expressed in its lowest terms as (\frac{p}{q}), must have (p) as a factor of the constant term (in our case, (d)) and (q) as a factor of the leading coefficient (in our case, (a)).

For example, consider the cubic equation (x^3 - 6x^2 + 11x - 6 = 0). The factors of the constant term (-6) are (\pm1, \pm2, \pm3, \pm6), and the factors of the leading coefficient (1) are just (\pm1). Therefore, the possible rational roots are (\pm1, \pm2, \pm3, \pm6).

Step 2: Test for Rational Roots

Next, we test these possible rational roots by substituting them into the equation to see if any of them satisfy the equation, meaning the result is zero. This process can be tedious but is a systematic way to find a starting point for factorisation.

Using our example (x^3 - 6x^2 + 11x - 6 = 0), let’s test (x = 1): [1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0]

So, (x = 1) is a root, which means (x - 1) is a factor.

Step 3: Perform Synthetic Division or Polynomial Long Division

Once a root is found, we can use synthetic division or polynomial long division to divide the cubic polynomial by the linear factor we’ve identified. This process will reduce the cubic equation to a quadratic equation, which we can then attempt to factorise further or use the quadratic formula on.

For (x^3 - 6x^2 + 11x - 6 = 0), dividing by (x - 1) gives us: [x^2 - 5x + 6 = 0]

Step 4: Factorise the Quadratic Equation

Now, we attempt to factorise the quadratic equation (x^2 - 5x + 6 = 0). We look for two numbers that multiply to (6) and add to (-5). These numbers are (-2) and (-3) because (-2 \times -3 = 6) and (-2 + (-3) = -5).

So, the quadratic equation factors into ((x - 2)(x - 3) = 0).

Step 5: Write the Complete Factorisation

Combining all our steps, the complete factorisation of the original cubic equation (x^3 - 6x^2 + 11x - 6 = 0) is: [(x - 1)(x - 2)(x - 3) = 0]

This means the roots of the equation are (x = 1), (x = 2), and (x = 3).

Conclusion

Factorising a cubic equation can be straightforward if we start by identifying and testing for rational roots. Once we find a root, we can use division to reduce the cubic to a quadratic, which is typically easier to factorise or solve using the quadratic formula. Remember, not all cubic equations will factor nicely, and some may require more advanced techniques, such as Cardano’s Formula for reduced cubic equations. However, for many educational and practical purposes, the method outlined here will suffice to factorise a cubic equation easily.