Question

How do you factor `c^2 + 6c + 9`?

Answer 1

See a solution process below:

Explanation:

Because the `x^2` coefficient is `1` we know the coefficient for the `c` terms in the factor will also be `1`:

`(c )(c )`

Because the constant is a positive and the coefficient for the `x` term is a positive we know the sign for the constants in the factors will both be positive because a positive plus a positive is a negative and a positive times a positive is a positive:

`(c + )(c + )`

Now we need to determine the factors which multiply to 9 and also add to 6:

`1 xx 9 = 9`; `1 + 9 = 10` <- this is not the factor

`3 xx 3 = 9`; `3 + 3 = 6` <- this IS the factor

`(c + 3)(c + 3)`

Or

`(c + 3)^2`

Answer 2

`(c+3)^2`

Explanation:

First.... look at the first term `c^2`
Now.... `c^2` can either be

• `(cxx x)(c xx y)`
• `(c^2 xx x)(1 xx y)` Now... notice that if it was the second possibility... there would be another `c^2` as: `(c^2 xx x)(1 xx y)=c^2+c^2y+x+xy` So.. possibility eliminated Which leaves us with only one possibility: `(c xx x)(c xx y)` Now.... what is `x` and `y`? Notice that both `6c` and `9` are multiples of `3` and also notice that `3+3=6` and `3xx3=9` Problem solved!! `(c+3)(c+3)` `(c+3)^2`