Factorising Harder Quadratics: Tricks You Need to Know

Harder Factorise

As an experienced mathematics tutor, I’ve seen many students struggle with factorising harder quadratics. The process might seem intimidating at first. However, there are several tricks and techniques you can use to simplify the task. These will help you solve even the most complex quadratic equations with confidence. In this article, we’ll explore some of these methods. Additionally, we’ll work through a series of examples to help you master factorising harder quadratics.

Understanding the Basics

Before we dive into the more advanced techniques, it’s crucial to have a solid grasp of the fundamentals. Specifically, a quadratic expression is an algebraic expression in the form of $ax^2 + bx + c$. Here, $a$, $b$, and $c$ are constants, and $a \neq 0$.

The goal of factorising a quadratic is to rewrite the expression as a product of two linear factors. This is typically in the form $(px + q)(rx + s)$, where $p$, $q$, $r$, and $s$ are constants. In other words, we aim to find the factors that, when multiplied together, result in the original quadratic expression.

Techniques for Factorising Harder Quadratics

When faced with harder quadratics, there are several techniques you can use to factorise them successfully. These include:

  1. Grouping
  2. Difference of squares
  3. Sum or difference of cubes
  4. Completing the square
  5. Factorise by substitution


Firstly, grouping is a useful technique when the quadratic expression has four terms. By grouping the terms in pairs and factoring out common factors, you can often simplify the expression. Subsequently, you can factorise it further. As a result, this technique can make the factorisation process more manageable. Moreover, it helps to break down the problem into smaller, more digestible parts. Additionally, grouping can be used together with other factorisation techniques for even greater effectiveness.

Difference of Squares

Secondly, when a quadratic expression is in the form of $a^2-b^2$, it can be factorised as $(a + b)(a-b)$. This technique is known as the difference of squares. Moreover, it can be applied to more complex expressions as well. Therefore, it’s essential to recognise this pattern when factorising harder quadratics. Furthermore, mastering this technique will save you valuable time and effort in the long run. Additionally, the difference of squares can be used to simplify expressions before applying other factorisation methods.

Sum or Difference of Cubes

Thirdly, expressions in the form of $a^3 + b^3$ or $a^3 – b^3$ can be factorised using the sum or difference of cubes formulas:

  • $a^3 + b^3 = (a + b)(a^2-ab + b^2)$
  • $a^3-b^3 = (a-b)(a^2 + ab + b^2)$

Consequently, memorizing these formulas can save you time and effort when factorising harder quadratics. Additionally, practising these formulas will help you recognize these patterns more easily in the future. Moreover, the sum and difference of cubes can be combined with other techniques to factorise even more complex expressions.

Completing the Square

Furthermore, when a quadratic expression cannot be easily factorised using other techniques, completing the square can be a valuable method. This involves rewriting the quadratic in the form of $(x + p)^2 + q$. Consequently, it can then be solved using the square root method. As a result, this technique can be particularly useful when dealing with quadratics that have no obvious factors. Moreover, it’s a powerful tool to have in your mathematical toolkit. Additionally, completing the square can be used to solve quadratic equations and to find the vertex of a parabola.

Factorise by Substitution

Lastly, in some cases, factorising by substitution can be an effective approach. This technique involves substituting a complex expression with a single variable, then factoring the resulting expression. Afterwards, you substitute the original expression back in. Thus, this method can simplify the factorisation process for more complicated quadratics. Furthermore, it can help you break down the problem into more manageable steps. Additionally, factorising by substitution can be used together with other techniques to tackle even the most challenging quadratic expressions.


Now that we’ve covered some of the key techniques for factorising harder quadratics, let’s work through a series of examples. These will help solidify your understanding. Additionally, these examples will demonstrate how to apply the techniques we’ve discussed. Moreover, they’ll provide you with valuable practice in factorising harder quadratics. Therefore, take your time to work through each example carefully. Furthermore, try to identify which technique(s) would be most appropriate for each problem before attempting to solve it.

  1. Factorise $( x^2-3x )^2-2x^2 + 6x-8$.
  2. Factorise $( x-1 )( x-3 )( x+2 )( x+4 ) + 24$.
  3. Factorise $x^4-5x^2 +4$.
  4. Factorise $x^4 + x^2 + 1$.
  5. Factorise $x^4-6x^2y^2 + y^4$.
  6. Factorise $x^3 + 2x^2y^2-x-2y$.
  7. Factorise $x^3 + x^2z + xz^2-y^3-y^2z-yz^2$.
  8. Factorise $( x^2-8x + 12 )( x^2-7x + 12 )-6x^2$.

Practice Makes Perfect

As with any mathematical skill, practice is crucial to mastering the art of factorising harder quadratics. Therefore, regularly work on a variety of problems, applying the techniques discussed above. Furthermore, don’t hesitate to seek help from your teacher or tutor when needed. By doing so, you’ll gradually build your confidence and proficiency in factorising harder quadratics. Moreover, you’ll develop a deeper understanding of the underlying concepts.

Additionally, as you practice, you’ll start to recognize patterns and develop an intuition for which techniques to apply in different situations. As a result, this will make the process of factorising harder quadratics more efficient and less daunting.

In conclusion, by understanding the basics and utilising the various techniques for factorising harder quadratics, you’ll be well-equipped to tackle even the most challenging quadratic expressions with ease.

Moreover, with perseverance and practice, you can become proficient at factorising harder quadratics in no time! So, keep working hard and don’t give up – success is just around the corner.

Furthermore, remember that every problem you solve brings you one step closer to mastering this essential mathematical skill. Lastly, always approach factorising harder quadratics with a positive attitude and an open mind, and you’ll be sure to succeed!

Consequently, you’ll find that factorising harder quadratics becomes second nature to you. Additionally, you’ll be able to apply these skills to a wide range of mathematical problems.

Question 1

Factorise \( (x^2-3x)^2-2x^2 + 6x-8 \).

\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
&= (x^2-3x)^2-2(x^2-3x)-8 \\
&= A^2-2A-8 &\color{red} \text{let } A = x^2-3x \\
&= (A + 2)(A-4) \\
&= (x^2-3x + 2)(x^2-3x-4) \\ &= (x-1)(x+2)(x-4)(x+1)
\end{aligned} \)

Question 2

Factorise \( (x-1)(x-3)(x+2)(x+4) + 24 \).

\( \begin{aligned} \require{AMSsymbols} \displaystyle
&= (x-1)(x+2) \times (x-3)(x+4) + 24 \\
&= (x^2 +x-2)(x^2 + x-12) + 24 \\
&= (A-2)(A-12) + 24 &\color{red} \text{let } A = x^2 + x \\
&= A^2-14A + 48 \\
&= (A-6)(A-8) \\
&= (x^2 + x-6)(x^2 + x-8) \\
&= (x-2)(x+3)(x^2 + x-8)
\end{aligned} \)

Question 3

Factorise \( x^4-5x^2 +4 \).

\( \begin{aligned} \displaystyle
&= (x^2-1)(x^2-4) \\
&= (x+1)(x-1)(x+2)(x-2)
\end{aligned} \)

Question 4

Factorise \( x^4 + x^2 + 1 \).

\( \begin{aligned} \displaystyle
&= x^4 + 2x^2 + 1-x^2 \\
&= (x^2 + 1)^2-x^2 \\
&= (x^2 + 1 + x)(x^2 + 1-x)
\end{aligned} \)

Question 5

Factorise \( x^4-6x^2y^2 + y^4 \).

\( \begin{aligned} \displaystyle
&= x^4-2x^2y^2 + y^4-4x^2y^2 \\
&= (x^2-y^2)^2-(2xy)^2 \\
&= (x^2-y^2 + 2xy)(x^2-y^2-2xy)
\end{aligned} \)

Question 6

Factorise \( x^3 + 2x^2y^2-x-2y \).

\( \begin{aligned} \displaystyle
&= 2(x^2-1)y + x(x^2-1) \\
&= (x^2-1)(2y + x) \\
&= (x+1)(x-1)(2y+x)
\end{aligned} \)

Question 7

Factorise \( x^3 + x^2z + xz^2-y^3-y^2z-yz^2 \).

\( \begin{aligned} \displaystyle
&= (x-y)z^2 + (x^2-y^2)z + x^3-y^3 \\
&= (x-y)z^2 + (x-y)(x + y)z + (x-y)(x^2 + xy + y^2) \\
&= (x-y)\big[z^2 + (x + y)z + x^2 + xy + y^2\big] \\
&= (x-y)(z^2 + xz + yz + x^2 + xy + y^2) \\
&= (x-y)(x^2 + y^2 + z^2 + xy + yz + zx)
\end{aligned} \)

Question 8

Factorise \( (x^2-8x + 12)(x^2-7x + 12)-6x^2 \).

\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
&= (A-8x)(A-7x)-6x^2 &\color{red} \text{let } A = x^2 + 12 \\
&= A^2-15Ax + 56x^2-6x^2 \\
&= A^2-15Ax + 50x^2 \\
&= (A-10x)(A-5x) \\
&= (x^2-5x + 12)(x^2-10x + 12)
\end{aligned} \)

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