How would you factor this problem? I keep getting stuck.
x^2+2xy+2x+y^2+2y-8
x^2+2xy+2x+y^2+2y-8 can be rewritten
x^2+2xy+y^2 + 2x+2y-8
=(x+y)^2 + 2(x+y-4) OR
= [(x+y)(x+y+2] -8)
Thanks so much. :)
You're wecome. Thanks for using Jiskha!
To factor the equation x^2 + 2xy + 2x + y^2 + 2y - 8, we can use a method called grouping. Here's how you can approach it step by step:
Step 1: Group the terms pairwise and look for common factors.
Group the terms with similar variables:
(x^2 + 2xy + 2x) + (y^2 + 2y) - 8
Step 2: Factor out any common terms within each group.
In the first group, we can factor out x:
x(x + 2y + 2) + (y^2 + 2y) - 8
Step 3: Look for common factors between all the groups.
In this case, we don't have any common factors among all the groups.
Step 4: Try to rearrange the equation to create a perfect square trinomial term.
To create a perfect square trinomial, we look for a term in a binomial form that, when squared, will give us the desired result. In this case, we want a term that would complete the square for x and y.
Step 5: Add and subtract the square of half the coefficient of the middle term within the binomial.
Let's focus on the x terms: x(x + 2y + 2). Half of the coefficient of 2y is y, and the square of y is y^2. Similarly, half of the coefficient of 2 is 1, and the square of 1 is 1^2. So we can add and subtract these terms within the equation:
x(x + 2y + 2 + y^2 - y^2 - 1) + (y^2 + 2y) - 8
Step 6: Rearrange the equation by the grouping.
Now, let's rearrange the equation so that we group the perfect square terms together:
(x(x + 2y + 2 - 1) + (y^2 + 2y - y^2)) + 2y - 8
Step 7: Simplify and combine like terms.
(x(x + 2y + 1) + 2y) + 2y - 8
Step 8: Factor common terms in each group and further simplify.
x(x + 2y + 1) + 2y + 2y - 8
x(x + 2y + 1) + 4y - 8
And there you have it! The factored form of the equation is x(x + 2y + 1) + 4y - 8.