# Explain how to factor a general expression x^2 + bx + c.

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### 2 Answers

To factor a general quadratic expression `x^2+bx+c` we will proceed as under -

Let `(x-alpha)` and `(x-beta)` are the factors of the above expression.

Then `x^2+bx+c=(x-alpha)(x-beta)`

or, `x^2+bx+c=x^2-(alpha+beta)x+alphabeta`

Comparing coefficients of powers of `x` from both sides we get

`alpha+beta=b` (1) and `alphabeta=c` (2).

Now `(alpha-beta)^2=(alpha+beta)^2-4alphabeta`

or, `(alpha-beta)^2=b^2-4c` (Using equations (1) and (2))

or, `(alpha-beta)=+-sqrt(b^2-4c)` (3)

Solving (1) and (3) we get (by taking the positive sign in (3))

`alpha=(-b+sqrt(b^2-4c))/2` ,

`beta=(-b-sqrt(b^2-4c))/2` .

If we take negative sin we get the same values provided the role of `alpha ` and `beta ` are changed.

So the factors are

`x-{(-b+sqrt(b^2-4c))/2}` and `x-{(-b-sqrt(b^2-4c))/2}` .

For a general expression x^2 + bx + c, it is not certain that it can be factored as that depends on what the values of b and c are.

One way of determining the factors is using the formula for the roots of a quadratic equation. The roots of x^2 + bx + c = 0 are given by the formula `(-b+-sqrt(b^2 - 4c))/2`

The factored form of x^2 + bx + c is:

`(x - (-b+sqrt(b^2 - 4c))/2)(x - (-b-sqrt(b^2 - 4c))/2)`

If the term `sqrt(b^2 - 4c)` yields an integer, the expression given was one that could be factored.

**Any expression x^2 + bx + c can be factored as **`(x - (-b+sqrt(b^2 - 4c))/2)(x - (-b-sqrt(b^2 - 4c))/2)`