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**x ^{2}+x-6=y**

Is the equation we will use to demonstrate how to factorise quadratics.

The first step involves using the basic shape of all quadratic factorisation:

**ax ^{2}+bx+c=y**

**x ^{2}+x-6=y**

**(Cx+A)(Dx+B)=y**

We must realize certain equalities that appear between the different expressions of this equation.

1.

Cx*Dx=ax^{2}

C*Dx^{2}=ax^{2}

Cancelling x^{2}

C*D=a

2.

A*B=c

3.

Cx*B+Dx*A=bx

B*Cx+A*Dx=bx

Cancelling x

B*C+A*D=b

This rigid layout can be used to factorise quadratics, but quadratics are all about pattern recognition and a small amount of practice goes a long way.

**x ^{2}+x-6=y**

**ax ^{2}+bx+c=y**

1. As our quadratic has no number multiplying on x^2 the first step of the solution is simple, we know that both C and D are equal to 1 as 1 only has one factor.

C*D=a

C*D=1

1*1=1

2. This is where paths in the solution diverge, as c in our equation, -6, has a number of factors

Those factors are:

+3*-2=-6

-3*+2=-6

+1*-6=-6

-1*+6=-6

A*B=-6

So we know the A and B are one of these factor pairs.

3. B*C+A*D=b

From step 1 in our solution, we know that both C and D are equal to 1. Meaning we can simplify our equation:

A+B=b

A+B=1

Now, from the factors we found in step 2, we must select a pair thats sum equals 1.

+3-2=1, so we know that A=+3 and B=-2 (it is arbritrary which number is assigned to each letter as the rest of the equation is the same).

**ax ^{2}+bx+c=y**

**x ^{2}+x-6=y**

**(Cx+A)(Dx+B)=y**

C=1

D=1

A=+3

B=-2

x^{2}+x-6=y

(1x+3)(1x-2)=y

Finally, checking our answer:

1x*1x+3x-2x-6=y

x^{2}+x-6=y

Following a rigid method is not recomended for solving quadratics, remember steps and the equalities that must occur, and practice, are the most important things.