Factorising

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Starting with the simple case (explaining with example)

We know that if there's a common factor in every term of an expression then we can factorise it out; \[8x + 4 = 4(2x+1)\] Similarly, if that factor is \(x\) then we get something like \[x^2+3x = x(x+3)\] So continuing with this example; \[\begin{align} x^2+3x&=0 \\[3pt]x(x+3)&=0 \end{align}\] \[x = 0\hspace{6 pt}\text{or }-3\]

"But why is the root the negative of the value in the bracket?"

We know that if the product of two numbers is \(0\) then at least one of them is \(0\). For the equation \(x(x+3)=0\) the first possibility is obvious, if \(x=0\) then \[0\times(0+3)=0\] We can also see that if \(x=-3\) then \[(-3)(-3+3)=(-3)\times0=0\]

The Usual Case (explaining with example)

More often, there will be a term which you cannot factor \(x\) from, such as \[x^2+5x+6=0\] So we have to try and factorise it into something of the form \[(x+a)(x+b)=0\] where \(a\) and \(b\) are real numbers and \(x\) is our variable. Let's take a look at how this would expand \[\begin{align} (x+a)(x+b) &= x^2 + ax + bx + ab \\[3pt]&= x^2 + (a+b)x + ab \end{align}\] Looking at the final form of this expansion, we can get an idea of how we're going to factorise our example! The \(x\) term has a coefficient which is the sum of \(a\) and \(b\), while the non-\(x\) term is the product of \(a\) and \(b\).
So we must ask ourselves "What numbers multiply to make \(6\) and add to make \(5\)?". You may be able to work out in your head what the answer to this example is. However, a more foolproof method is to look at factor pairs of \(6\), and which of those add to make \(5\).
Remember negative factors! \[\text{Factor pairs of }6\text{: }\{1,6\},\{2,3\},\{-1,-6\},\{-2,-3\}\] \[2+3=5\] So our answer is \[x^2+5x+6 = (x+2)(x+3) = 0\] \[x = -2\hspace{6 pt}\text{or }-3\] We can check by expanding \[(x+2)(x+3) = x^2+2x+3x+6 = x^2+5x+6 = 0\]

Full worked example: positive values

\[x^2+8x+16=0\] \[\text{Factor pairs of }16\text{: }\] \[\{1,16\},\{2,8\},\{4,4\},\{-1,-16\},\{-2,-8\},\{-4,-4\}\] \[4+4=8\] \[(x+4)(x+4)=0\] \[x = -4\hspace{10 pt}\text{(repeated root)}\]

Full worked example: negative values

\[x^2-6x-7=0\] \[\text{Factor pairs of }-7\text{: }\{-1,7\},\{1,-7\}\] \[(-1)+7=6\] \[(x-1)(x+7)=0\] \[x = 1\hspace{6 pt}\text{or }-7\]

Full worked example: mixed values

\[x^2-11x+28=0\] \[\text{Factor pairs of }28\text{: }\] \[\{1,28\},\{2,14\},\{4,7\},\{-1,-28\},\{-2,-14\},\{-4,-7\}\] \[(-4)+(-7)=-11\] \[(x-4)(x-7)=0\] \[x = 4\hspace{6 pt}\text{or }7\]