Question

Solve by factoring:
y^2 + 2y = 8
can someone help me out here? D:

Answer 1

very easy i will do

Answer 2

bring all to one side:

y^2+2y-8=0
factorise to get (y+4)(y-2)=0

that means y + 4 = 0, or y - 2 = 0
y = -4,2

Answer 3

so then I would have to do y^2 + 2y - 8 = 8-8 and then y^2 + 2y - 8 = 0 how do I get it to (y + 4)(y - 2)=0?

Answer 4

bring all to one side:

y^2+2y-8=0
factorise to get (y+4)(y-2)=0

that means y + 4 = 0, or y - 2 = 0
y = -4,2

Answer 5

y^2 + 2y - 8. We split the middle term.

But first, what is the product of the first and last coefficients? (1 and -8)
-8

Now, what two numbers multiply to get -8 and add to get 2 (the second coefficient)
-2 and 4.

y^2 - 2y + 4y - 8 = 0

Now group in pairs.

y (y - 2) + 4 (y - 2) = 0
(y + 4) (y - 2) = 0