how to factorise cubics. The Beyond Maths blog has tutorial-style posts like this one on how to factorize a cubic that can help you out if you’re either getting ready for advanced math study or in the middle of the mayhem.
Working examples, diagrams, grids, and practice questions help bring the concepts and techniques described in this book to life so that you can learn by doing.
So, let’s dive in.
Three steps are required to factorize a cubic expression:
To discover a linear factor, please.
To get a quadratic, divide by the linear factor.
The quadratic must be factored.
It’s a continuation of the material covered in your GCSE years. Using the resource here, you can test your knowledge of the factor theorem and polynomial division and prepare for your GCSE exams. When you’re comfortable factoring cubes, give this difficult exercise a shot.
Step 1: Locate a linear factor
The form (x + b) or (ax + b) will appear whenever a cubic is factored into a linear form. To factor a cube, you must know its value of b. (and sometimes a).
Use the factor theorem to determine this factor. Given a function f(x), the factor theorem states that
If (x-b) is a factor, then f(b) = 0; likewise, if (ax-b) is a factor, then f(fracbax) = 0.
Consider the cube below, whose factor is (x + 2):
f(x) = 2x^3 + 11x^2 + 17x + 6
It is easy to demonstrate this by finding the value of f(-2); if (x + 2) is a factor, then f(-2) will be equal to 0.
A starting alignment of f(-2) would look like this: 2(-2)3 + 11(-2)2 + 17(-2) + 6 &= 2 times. -8 + 11 times 4 + 17 times (x + 2) is a factor because f(-2) = 0 and (x + 2) is a factor of (16, 44, -34) plus 6.
You were given a factor in that scenario. But what if you’re on your own and have to locate one?
Think about what occurs when a cube is made larger:
To simplify, we can write: (x + 2)(x + 3)(x – 5) = (x2 + 5x + 6)(x – 5) = x3 – 19x – 30
Multiplying the constant term in each bracket yields the constant term of -30.
Multiplying 2 by 3 by -5 yields -30.
Since the constant in the cubic expression is a factor of the constant in any linear factor, the two cannot be independent. In this way, you have a baseline from which to investigate other factors.