# How to Solve Quadratics by Factorising

The method of how to solve quadratics by factorising is now part of the foundational knowledge students aiming for higher exam grades are expected to have.   Here is an example of such a question.

Solve x2 + 7x – 18 = 0

In my experience of teaching and marking exam papers students often struggle with solving equations by factorisation.  Common incorrect attempts include trying to manipulate the equation using the balance method or using a method of trial and improvement.  At best, both attempts only lead to one solution and are therefore not awarded any marks as no creditable attempt to factorise has been made.

In this blog I show how to solve quadratics by factorising using scaffolded questions and mini whiteboards to check student’s understanding throughout a lesson.

At the start of the lesson students revise how to factorise a quadratic into two brackets.  The purpose of this is to remind the class that some quadratics can be written as a product of two brackets.  I encourage students who have forgotten how to factorise a quadratic to multiply out the brackets on the right.

When all the expressions are matched, they should look for a pattern between the numbers in each bracket and the numerical terms in the matching quadratic.  The pattern being the constant term is their product and the coefficient of x their sum.

## How to Solve Quadratics by Factorising

To solve a quadratic, we discuss if it can be represented as a product of two brackets. When each bracket equals zero the quadratic itself equals zero.  In this video I demonstrate the method further.

Having demonstrated a few examples I ask the class to solve x2 + 11x + 24 = 0 and show me their working on mini whiteboards.  It is pleasing that all students have correctly factorised the quadratic into (x + 3)(x + 8) = 0 but some present 3 and 8 as their solutions for x.

To feedback we discuss how each bracket must equal zero for the overall quadratic to be zero.  For example, x + 3 = 0, x = -3 and x + 8 = 0, x = -8.  I demonstrate these are the correct solutions by substituting the values -3 and -8 back into the original quadratic.

When working through the following questions on mini whiteboards students experience some difficulty in dealing with combinations of negative and positive factors of the constant term.

x2 + 3x – 4 = 0

x2 – 11x + 28 = 0

However, this is more to do with needing greater practice with factorisation than it does understanding how to solve a quadratic.

Solving quadratics with a zero term, such as:

x2 – 9 = 0

x2 – 3x = 0

also proved challenging for some students.  To address this, I ask the students to consider the value of the term that is missing.  In the case of x2 – 9 = 0 the b term is 0.  Therefore, we need two numbers that multiply to make -9 but have a sum of 0.  For x2 – 3x = 0 we need two numbers that have a product of 0 but a sum of -3.  About 25 minutes into the lesson students are ready to work independently through the questions below.

## Using Factorisation to Solve Real-Life Problems

In the final part of the lesson students apply their learning to model a real-life problem as shown.

To help the class set up the quadratic equation I encourage students to consider the height of the ball from the ground at time t = 0 and the value of t when the height is 96 m above the ground.

About 5 minutes later, when the class have had time to sketch out some ideas, I provide further prompts for those who are struggling.  These include how to make the quadratic equal to zero and to factorise out the 2 from 2t­2 – 28t – 96 = 0 to make t2 – 14t + 48 = 0.

Having factorised and solved the equation some of the students are not sure whether the time is 6 or 8 seconds.  My response is to sketch the graph of the changing height of the ball as a function of time.  The resulting parabola helps all students understand the ball passes 96 m travelling upwards after 6 seconds and again travelling downwards at 8 seconds.

## Next Steps

I recently taught this to a Higher GCSE class as the first lesson on solving quadratic equations.  The next lesson is to solve quadratics where the coefficient of x2 is no longer one.  If I had taught this to a Foundation class, the next lesson would be to model and solve real life problems through factorising quadratics.`

## Related Lessons ## Revising Quadratic and Linear Simultaneous Equations

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