Question

How do you factor `9 + 6a + a² - b² - 2bc - c²`?

Answer 1

`9+6a+a^2-b^2-2bc-c^2 = (a-b-c+3)(a+b+c+3)`

Explanation:

Notice that both of the following are perfect squares:

`9+6a+a^2 = (a+3)^2`

`b^2+2bc+c^2 = (b+c)^2`

The difference of squares identity can be written:

`A^2-B^2=(A-B)(A+B)`

So with `A=(a+3)` and `B=(b+c)` we find:

`9+6a+a^2-b^2-2bc-c^2 = (9+6a+a^2)-(b^2+2bc+c^2)`

`color(white)(9+6a+a^2-b^2-2bc-c^2) = (a+3)^2-(b+c)^2`

`color(white)(9+6a+a^2-b^2-2bc-c^2) = ((a+3)-(b+c))((a+3)+(b+c))`

`color(white)(9+6a+a^2-b^2-2bc-c^2) = (a-b-c+3)(a+b+c+3)`

Answer 2

`9+6a+a^2-b^2-2bc-c^2=(a+3+b+c)(a+3-b-c)`

Explanation:

`9+6a+a^2-b^2-2bc-c^2` can be written as

`a^2+6a+9-(b^2+2bc+c^2)`

or `[a^2+2xx3xxa+3^2]-{b^2+2xxbxxc+c^2]`

Using identity `x^2+2xy+y^2=(x+y)^2`, this becomes

`(a+3)^2-(b+c)^2`

and now using `x^2-y^2=(x+y)(x-y)`, this becomes

`(a+3+b+c)(a+3-b-c)`