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Firstly, make sure you agree that a quadratic equation is an equation of the form y = ax2 + bx + c where a,b,c are (real) constants (a is not 0), and x is the variable. (Note: the equation has an "x squared" term).
An example would be: y = x2 + x - 6.
We want to factorise this equation so we can find the values of x for which y = 0 (the points where the curve crosses the x-axis).
Secondly, we need to be aware that factorising a quadratic mens expressing the equations as a product of two brackets which each contain an x term.
So, in the form ax2 + bx + c = (dx + e)(fx + g)
Working with our example, y = x2 + x - 6, we first direct our attention to the constant term c, in this case c = -6.
If we expand the brackets, we get ax2 + bx + c =(dx+e)(fx + g) = dfx2 + (ef +dg)x + eg.
We should already be able to see that for our example where the coefficient of x2 is 1 that d,f = 1 so now we have a simpler equation:
(x+e)(x+g) = x2 + (e + g)x + eg. (1.1)
We can then use eg = c = -6, our constant term ie. the constant terms in the brackets multiple to make our origanal constant term.
To find e,h, we think of pairs of numbers which multiply to give -6 :
-1 x 6 // -2 x 3 // -3 x 2 // -6 x 1
So how do we decide which pair of number will give the correct equation?
Well we could
test each pair and multiply out the brackets until we get the right equation, but this could take some time if we have more than four options, so instead we'll take a shortcut:
See which pairs add to give the coefficient of x
From (1.1), we can equate e + g = b.
-1 + 6 = 5 // -2 + 3 = 1 // -3 + 2 = -1 // -6 + 1 = -5
In our example, b = 1, so we can tell that our constants, e,h are -2,+3 and our answer is: y = (x - 2)(x + 3).
Check! (x - 2)(x + 3) = x2 - 2x + 3x - 6 = x2 + x - 6
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