Question

I need help factoring

Answer 1

do ya know about the snow flake method?

Answer 2

no

Answer 3

i used to know how to do this but forgot

Answer 4

This is actually a circle I believe. The x terms factor to \[(x-2)^{2}\]and the y terms factor to \[(y+4)^{2}\]Set it equal to 16 which is the radius squared to get the equation of the circle as such:\[(x-2)^{2}+(y+4)^{2}=16\]

Answer 5

You do not need to factor this in the traditional sense; just put it into standard form for a circle.

Answer 6

can you show me step by step?

Answer 7

Sure, but there really isn't anything much to show except that you see that you have a polynomial in x added to a polynomial in y set equal to some number that just happens to be a perfect square.

Answer 8

These expressions are each quadratic and there are certain rules about quadratics. Like the fact that circles are always an x^2 term PLUS a y^2 term set equal to a perfect square. Then parabolas are either x^2 OR y^2 equal to y or x, respectively. Then hyperbolas are fractions x^2 MINUS y^2 or y^2 MINUS x^2 over a or b, respectively. And it goes on with ellipses also. So basically you need to recognize the fact that these are polynomials and because they are added to one another you will recognize that is a circle formula in expanded form. A quadratic equation will either be a circle or a hyperbola or a parabola or an ellipse. they all have different standard forms; you just need to learn how to recognize which is which. Really that is it

Answer 9

if you expand the (x-2)^2, you will get x^2-4x+4, and if you expand the (y+4)^2, you will get y^2+8y+16. See that? (x-2)^2 is a perfect square binomial, as is (y+4)^2

Answer 10

Also, the numbers inside the parenthesis with the x and the y indicate the center of the circle. This is a circle centered at (2,-4) with a radius of 4.

Answer 11

Thank you so much for yout help