 At school I was never taught why factorising quadratic equations is useful. Only when I did a Physics degree did I really realise how often solving a quadratic crops up in real science and engineering problems. The reason why we go through the pain of learning to factorise quadratics is so that we can solve them easily. Let's go through it step-by-step.

# What is solving an equation?

When we solve an equation we're finding the values of the variables which make the equation "true". So say we have y=0. The value of y for which that is true is zero. If y=3 then the equation is not true. So the solution is easy, it's y=0.

What's 3 times 0? What's 0 times 3? Zero, right? Now if I tell you that I have two numbers, call them x and y, and that their product is zero what does that tell you?

xy=0

It tells you that either x is zero or y is zero, or they are both zero.

What if the two numbers we're multiplying together are:

(x-3)(x+4)=0

We know, just as before, that either x-3=0 or that x+4=0. If the first is true then x=3 and if the second is true then x=-4.

That's why we factorise quadratics - it makes solving quadratic equations really easy. If we had used the unfactorised version of the quadratic it wouldn't have been easy at all.

$$x^2+x-12=0$$

Sometimes it helps to be able to see what's going on with a graph. If we plot the graph of x versus y the shape of a quadratic is called a parabola and it looks like this. You can see the solutions where the quadratic crosses the x-axis i.e. where y=0 marked with the dotted lines. The parabola cuts the x-axis in two places which is why there are two solutions to the equation. The first trick is to look at the sign of the number term. Remember the following:

• Positive number times positive number = positive number
• Negative number times negative number = positive number
• Positive number times negative number = negative number
• Negative number times positive number = negative number

Or to summarize those possibilities:

• If signs are different, product is negative
• If signs are the same, product is positive

Now the number term in a quadratic is the key to finding the solution. Let's look at some examples:

$$x^2-9x+20=0$$
• The sign of the number term (+20) is positive. So the signs of the numbers are either + and + or - and -. That means we need two numbers that add up to -9 where the signs are either both positive or both negative. The factors of twenty are
• 20 x 1, so solution would have -20 and -1 or +20 and +1
• 10 x 2, so solution would have -10 and -2 or +10 and +2
• 5 x 4, so solution would have -5 and -4 or +5 and +4
• The only factor pair that adds up to -9 is -5 and -4, so the solution is (x-4)(x-5).

Here's another

$$x^2+3x-18=0$$
• The sign of the number term (-18) is negative. So the signs of the numbers are either - and + or + and -. That means we need two numbers that add up to +3 where the signs are different. The factors of 18 are
• 18 x 1, so solution would have -18 and +1 or +18 and -1
• 9 x 2, so solution would have -9 and +2 or +9 and -2
• 6 x 3, so solution would have -6 and +3 or +6 and -3
• The only pair that adds up to +3 is +6 and -3, so the solution is (x+6)(x-3).

## Layout

I've seen two approaches to laying out your rough working for solving quadratics. The first is to write out the brackets with blanks and try filling them in, like this:

( x    )( x    )

Then you fill in your various attempts to solve the problem. You either work in pencil and rub out your attempts or you fill the page with your trial solutions.

The second way is to draw a cross then the numbers go on the right and the factors of x on the left. Both "work" so you've just got to find one that you're comfortable with.

## A bit harder

The examples so far have just had an x^2. If the equation is something like

$$6x^2+11x-10=0$$

this makes factorising more difficult because now there are lots of possibilities for the multiplier of x^2. Fortunately the trick for the sign of the number term still works. In this case the number is -10 so the numbers are either positive and negative or negative and positive. There are two possible factor combinations to consider:

• 10 x 1, so the solution would have (...x - 10)(...x+1) or (...x + 10)(...x - 1)
• 5 x 2, so the solution would have (...x - 5)(...x + 2) or (...x +5)(...x -2)

The added headache is we have to work out the factors for the x terms too. The factors would be 6 x 1 or 3 x 2. The upside is that we don't have to worry about the sign. But there are now sixteen possibilities:

• (6x-10)(x+1) or (6x+10)(x-1), neither possible as they'd give -4x or +4x
• (6x+1)(x-10) or (6x-1)(x+10), neither possible as they'd give -59x or +59x
• (6x-5)(x+2) or (6x+5)(x-2), neither possible as they'd give 7x or -7x
• (6x-2)(x+5) or (6x+2)(x-5), neither possible as they'd give 28x or -28x
• (3x-10)(2x+1) or (3x+10)(2x-1), neither possible as they'd give -17x or +17x
• (3x-1)(2x+10) or (3x+1)(2x-10), neither possible as they'd give 28x or -28x
• (3x-5)(2x+2) or (3x+5)(2x-2), neither possible as they'd give -4x or +4x
• (3x-2)(2x+5) or (3x+2)(2x-5), one possible as they'd give 11x (yes!) or -11x

You can see how tedious this becomes. I personally find the cross method faster for these more difficult quadratics. Here's the cross method applied to the example above. The cross visually lets you multiply the x coefficient by the appropriate number very quickly. The correct solution is in the top-right of the image. You can rapidly eliminate each pair of possibilities for the number terms by pairing them. For example (3x+10)(2x-1) and (3x-10)(2x+1) both produce either plus or minus 17x which doesn't work because it should give 11x so we can eliminate both possibilities. You can see the only pair of numbers that works is 2 and -5 and -2 and +5 because these give plus or minus 11x. ## Completing the square

If you look at the exercises below you can try a really useful technique for solving quadratics that sometimes crops up in exams. It's a pattern you should be able to spot very quickly. In the exercises I ask you to expand (x+1)^2, (x+2)^2 and so on, like so:

$$(x+1)^2=(x+1)(x+1)=x^2+x+x+1=x^2+2x+1$$
$$(x+2)^2=(x+2)(x+2)=x^2+2x+2x+4=x^2+4x+4$$
$$(x+3)^2=(x+3)(x+3)=x^2+3x+3x+9=x^2+6x+9$$
$$(x+4)^2=(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16$$

The pattern you should recognize is: x^2+number * x+another number, particularly if the coefficient for x is an even number. That's because solving the quadratic becomes a lot easier if you can "complete the square". This means you rewrite the quadratic as (x+n)^2 where n is half the x coefficient then add some number to make the number term match correctly. So say we have x^2+4x+7. Let's complete the square.

$$x^2+8x+7=(x+4)^2-9$$

Why complete the square? Well, it lets us do two things really easily. Firstly we can solve the quadratic. If x^2+8x+7=0 then we can take the number to the right hand side, take the square root of both sides (remembering that the square root of a number can be positive or negative) and we have the solution! In the worksheet we actually derive the quadratic formula by completing the square.

$$(x+4)^2-9=0$$
We add nine to both sides to move the number to the right hand side.
$$(x+4)^2=9$$
Then we get rid of the square by taking the square root of both sides.
$$x+4=\pm 9$$
And finally we move the 4 over to the right hand side by subtracting 4 from both sides
$$x=-4 \pm 9$$
So the solutions are x=-4-9=-13 and x=-4+9=5.

The other thing that we can do easily is work out the x-value of the minimum of the parabola. When we have a number, say x, and we square it we throw away the minus sign. The smallest value that x^2 can have is when x is zero. The same is true of (x-3)^2, its lowest value occurs when x-3=0, which is when x=3. If we move the parabola up and down by adding or subtracting a number e.g. (x-3)^2 - 6 or (x-3)^2 + 5 the minimum value doesn't move. So once we have completed the square we know the minimum value. In the example above (x+4)^2 is at a minimum when x=-4.

## How many solutions to a quadratic?

The quadratic formula gives us a very useful tool for finding out whether a quadratic has a solution. There are three possibilities:

• Two Solutions: the parabola cuts the x-axis in which case there are two "solutions" or crossing points.
• One Solution: the parabola just touches the x-axis so there is one solution.
• No Solutions: the parabola hovers above the x-axis so there are no solutions at all! The magic formula that lets us work out the number of solutions is called the discriminant. It is

$$b^2-4ac$$

Remember that we take the square root of this term in the quadratic formula

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

We can't take the square root of a negative number, so when b^2-4ac is negative the quadratic has no solutions. The three possibilities are:

• Discriminant < 0: no solutions
• Discriminant = 0: one solution
• Discriminant > 0: two solutions