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Factorization

Quadratic Functions

Generally speaking, factorisation is the reverse of multiplying out. One important factorisation process is the reverse of multiplications such as this:

(x − 5)(x + 3) = x² − 2x − 15.

Here we have muliplied out two linear factors to obtain a quadratic expression by using the distributive law.

To factorise a quadratic we have to go from an expression such as x² − 2x − 15 to the linear factors (x − 5)(x + 3) which generate it when multiplied out. How can we do this?

The clue lies in the solutions of the equation x² − 2x − 15 = 0 (called a quadratic equation). If we factorise the quadratic, the equation can be written as (x − 5)(x + 3) = 0. But a product of two factors can only be equal to zero if one or the other factor is equal to zero. So for the equation to hold, either x − 5 must be zero or x + 3 must be zero. Therefore the two possible solutions of the equation are x = 5 and x = −3. Looking at it the other way around, if we knew the solutions of the equation, we could find the factors: they just take the form (x − (solution)). In this example, the two solutions are 5 and −3, and so the two factors are (x − 5) and (x − (−3)) which simplifies to (x − 5) and (x + 3).

So it looks as though if we could solve quadratic equations, we could factorise quadratics. But there is a formula for solving quadratic equations which you have probably seen before: it says that

The solutions of the equation  Ax2 + Bx + C = 0 are given by the formula

For example, the quadratic x² − 2x − 15 has A = 1, B = −2 and C = −15. So according to the formula the solutions of the equation x² − 2x − 15 = 0 are:

Still more exercises for practice:

The third and final complication occurs for quadratic expressions where A is not equal to 1.

Now you are fully armed to tackle any quadratic expression. Try these quadratic exercises.

Exercise

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