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### 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