Factoring quadratic when 'a' is not 1

Factoring a quadratic expression when 'a' is not 1 has the same basic procedure as described in splitting the middle term of a quadratic expression. Let us discuss an example:

`3x^2 + 5x + 2`

In the above quadratic expression, 'a' is not 1 but 3, 'b' is 5 and 'c' is 2. Product of 'a' and 'c' is 6. So, simply, you have to find two numbers which multiply to give 6 and add up to give b, that is 5. That is,

|__| * |__| = 6
|__| + |__| = 5

can you think of two such numbers? Hmm..

I got it! two times three is six and the sum of two and three is five! Now put `2x + 3x` in place of `5x` in the quadratic expression:

`3x^2 + 2x + 3x + 2`

Rearrange so you have the two three's together and the two two's together

`3x^2 + 3x + 2x + 2`

Factor out `3x` from the first two terms and `2` from the next two terms

`3x(x + 1) + 2(x + 1)`

Factor out `(x + 1)` from the quadratic,

`(x + 1)(3x + 2)`

Generalization

So what do we do when 'a' is not 1? Is there any special procedure to factor a quadratic when 'a' is not equal to 1?

The procedure is same, as we saw above, but when a is not 1 in a quadratic expression, in order to factor it you must remember to

• Multiply `a` with `c` to get their product
• Find two numbers whose product is equal to the above product and whose sum is equal to `b` in the quadratic expression