Question

How would I solve (x-3)^2 + 1

Answer 1

I would think it would be (x^2 + 9) +1, but in my answer key it says (x^2 + 6x + 9) +1
--> x^2 + 6x +10.

But I don't understand how they got that.

Answer 2

(x-3)^2 means (x-3)*(x-3)

(just like x^2 means x*x)

if we had, for example: A(x-3) and distribute the A, we would get
A*x - 3*A or
Ax - 3A

this works even if A is "complicated". Example: if A were (x-3) it would still be true, and we would have
(x-3)x - 3(x-3)

if we rewrite (x-3)x as x(x-3) (when multiplying, we can change the order), we have

(x-3)*(x-3) = x(x-3) - 3(x-3)

now distribute again: x(x-3) is x*x - 3*x or x^2 -3x
-3(x-3) is -3x +9

we get
(x-3)*(x-3) = x^2 -3x -3x +9

finally, -3 x's take away 3 more x's is simplified to -6x
(x-3)^2 = x^2 -6x +9

Answer 3

Thank you so much! i understand it now.

Answer 4

See
http://www.khanacademy.org/math/algebra/multiplying-factoring-expression/multiplying-binomials/v/multiplying-binomials

for how to do the multiply a bit faster. (people use FOIL to remember the details)

Answer 5

thanks!