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



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rakesh05's profile pic

Posted on

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}` .

justaguide's profile pic

Posted on

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)`

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