# Find the value of the polynomial 3xy+7x^2−3x^2y+3y^2x−2x^2−2xy+4x^2y−2y^2x, when x=1 and y=−2

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Answer:[tex]5[/tex]First Method:Substitute the given values in for [tex]x[/tex] and [tex]y[/tex].Solve.This isn't the only method, but it seems the least likely to result in an incorrect answer.If [tex]x=1[/tex] and [tex]y=-2[/tex], find the solution to:[tex]3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x[/tex] Substitute [tex]1[/tex] for [tex]x[/tex] and [tex]-2[/tex] for [tex]y[/tex].[tex]3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x\\3(1)(-2)+7(1)^2-3(1)^2(-2)+3(-2)^2(1)-2(1)^2-2(1)(-2)+4(1)^2(-2)-2(-2)^2(1)[/tex]Wow. That's a lot. Let's try simplifying things. I'd start by multiplying all our [tex]1[/tex]'s because any number multiplied by [tex]1[/tex] is that very same number, so we can easily reduce the length of our equation without much worry for messing up our math in the process.[tex]3(-2)+7-3(-2)+3(-2)^2-2-2(-2)+4(-2)-2(-2)^2[/tex]It still looks like torture trying to figure out this equation, but let's take it in chunks instead of all at once.[tex]3(-2)+7-3(-2)+3(-2)^2-2-2(-2)+4(-2)-2(-2)^2[/tex]First, I'm going to deal with all the exponents, and turn them into integers.[tex]3(-2)+7-3(-2)+3(4)-2-2(-2)+4(-2)-2(4)[/tex]Notice that things are starting to look less complicated. We are taking our math in chunks. This will make it less likely for us to mess up, although it is more time consuming.Now, let's do multiplication. Any number within a parentheses is going to be multiplied by the number next to it. Remember that this only applies when there is no sign between them. I'm going to go one at a time.[tex]3(-2)+7-3(-2)+3(4)-2-2(-2)+4(-2)-2(4)\\-6+7-3(-2)+3(4)-2-2(-2)+4(-2)-2(4)\\-6+7+6+3(4)-2-2(-2)+4(-2)-2(4)\\-6+7+6+12-2-2(-2)+4(-2)-2(4)\\-6+7+6+12-2+4+4(-2)-2(4)\\-6+7+6+12-2+4-8-2(4)\\-6+7+6+12-2+4-8-8\\[/tex]This is a lot of steps, but hopefully it doesn't looks as bad after seeing that I just did multiplication over and over. I'll go over another method after this that may make things less complicated.Now, we add these together. It doesn't matter what order you do this in.[tex]-6+7+6+12-2+4-8-8\\7+4+6+12-6-2-8-8\\11+18-8-16\\29-24\\5[/tex]Our final answer is [tex]5[/tex]. Here's another method:If [tex]x=1[/tex] and [tex]y=-2[/tex], find the solution to:[tex]3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x[/tex]Because we know that any number multiplied by [tex]1[/tex] will be that same number, let's just substitute in [tex]x[/tex] for now.[tex]3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x\\3(1)y+7(1)^2-3(1)^2y+3y^2(1)-2(1)^2-2(1)y+4(1)^2y-2y^2(1)\\[/tex]Now, just like before, we can multiply [tex]1[/tex] without doing complicated math. Just basically get rid of it even when it's squared because [tex]1^2=1*1=1[/tex].[tex]3(1)y+7(1)^2-3(1)^2y+3y^2(1)-2(1)^2-2(1)y+4(1)^2y-2y^2(1)\\3y+7-3y+3y^2-2-2y+4y-2y^2\\3y^2-2y^2+3y-3y-2y+4y+5\\y^2+2y+5[/tex]Looks a LOT less complicated. Substitute [tex]-2[/tex] in for [tex]y[/tex] and get your final answer.[tex]y^2+2y+5\\(-2)^2+2(-2)+5\\4-4+5\\5[/tex]