Factorising Quadratics
Factorising Quadratics
In this video we’re going to take our factorising to the next level with some Quadratic Factorising. But first, we need to discuss exactly what a quadratic is.
A quadratic is essentially when the equation looks something like this – \(ax^2 + bx + c = 0\) . The end goal of factorising these equations is to have 2 brackets that when you expand, will give you the initial quadratic formula.
Some questions will ask you to ‘solve’ these equations but I’ll show you how to do that a bit later in the video.
In the above, life is much easier if \(a =1\) as it means that our brackets will look something like this: \((x +c)(x+c)\). When \(a\) doesn’t equal 1, then it gets a little more tricky.
First, let’s have a look at how we should go about factorising quadratics when our \(a = 1\):
Factorising Quadratics when \(a=1\)
- Most questions will give you the formula in the format of \(a^2+bx+c\), however, if this isn’t the case, rearrange the equation to get it into this format.
- Create our brackets containing \(x\) i.e. \((x\space\space\space\space\space\space\space\space)(x\space\space\space\space\space\space\space\space)\).
- Here’s where the real thinking comes in – we need to find 2 numbers that when multiplied, will give us the same number as \(c\) but when added together, will give us the same number as \(b\).
- Finally, we need to add a \(+\) or \(-\) into each bracket in a way that if they were to be expanded, we would get our original formula.
Let’s have a quick look at an example:
\(x^2 – 2x – 3\)
So let’s run through our steps and see what we get as our answer:
- We’ve got it in our desired format, we can see that \(a = 1, b = -2, c=-3\)
- Let’s write in our brackets: \((x\space\space\space\space\space\space\space\space)(x\space\space\space\space\space\space\space\space)\)
- We want to find 2 numbers that multiply to give -3 and add to give -2. That would be +1 and -3!
- We then add these into our brackets to give a final answer of \((x+1)(x-3)\).
- BONUS STEP – if our question wants us to ‘solve’ the equation, they want a value to be assigned to \(x\) – so all we do is make each bracket = 0
\((x+1) = 0\) —> minus 1 from each side;
\(x = -1\)
\((x-3) = 0\) —> add 3 to each side;
\(x=3\)
So the solution to ‘solve \(x^2 -2x -3\)’ would be \(x=-1, x=3\).
For some comfort, you can sub in either value into the equation and you’ll get zero both times.
We can do a final check to make sure we’re right and multiply these brackets out to see if it gives us our original equation:
We would get \(x^2-3x+x-3\) and when we collect our terms together, our final equation is \(x^2-2x-3\) which means we’re correct.
Factorising Quadratics when \(a\) does not = 1
The method for this is similar to the quadratics where \(a = 1\), however there are some differences.
- Get into desired format – same as when \(a = 1\)
- Write out the initial brackets, however you need to put the value of \(a\) into a bracket – e.g. if we have \(2x^2 – 5x – 3,\) we would need the brackets to be \((2x\space\space\space\space\space\space\space)(x\space\space\space\space\space\space\space\space)\).
- We need to find 2 numbers that multiply to give \(c\), ignoring the sign of \(c\).
- This is where it gets a little tricky. Once you’ve found the numbers that multiply to equal \(c\), you need to try putting these numbers into the brackets and seeing what they multiply out to be.
It’s really important here, that you try the numbers in each bracket because if you put a number in with the \(2x\), it will give a different equation than if you were to put it in the bracket with the single \(x\).
Let’s use our above example of \(2x^2-5x-3\), we know the brackets will be \((2x\space\space\space\space\space\space\space)(x\space\space\space\space\space\space\space\space)\).
Now what multiplies to equal -3? Well it can only be either -1 & +3, or -3 & +1.
So let’s try subbing these in for our answer:
\((2x-3)(x+1)\) —> if we expand the brackets, we would get \(2x^2+2x-3x-3\) and when we collect our terms, we get \(2x^2-x-3\) which is not what we want.
Now lets give \((2x+1)(x-3)\) a go.
When we expand the brackets out, we get \(2x^2 -6x + x-3\), when we collect our terms, we get \(2x^2 -5x-3\) which is our original equation!
So we now know our answer is \((2x+1)(x-3)\), what do we do if the question wants us to solve it? Make each bracket equal to zero, exactly the same as when \(a\) does = 1.
\((2x+1) = 0\) —> minus 1 from each side;
\(2x=-1\) —> divide both sides by 2;
\(x = -\frac{1}{2}\)
\((x-3)=0\) —> add 3 to both sides;
\(x = 3\)
Therefore, our answer is \(x=3,x=-\frac{1}{2}\).
Hopefully, you enjoyed the video. As always, if you have any questions, please don’t be afraid to ask them in the comments below and I’ll be sure to get back to you.
Just to make sure you’re happy – give the below practice worksheets a go!