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Distributive property: (2y^2 + 3y - 1)(y^2 + 4y + 5)



that's all i got down so far

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Jaz S. | Math Made ManageableMath Made Manageable
4.5 4.5 (4 lesson ratings) (4)
Maralyn, You would multiply each term in the first polynomial (2y^2+3y-1) by every term in the second (y^2+4y+5). So what you have down so far is correct! For example, if I replaced the terms with letters - (2y^2+3y-1)=(A+B-C)and (y^2+4y+5)=(x+y+z) and used the distributive property you would get something like this: A(x+y+z)+B(x+y+z)-C(x+y+z) or Ax+Ay+Az+Bx+By+Bz-Cx-Cy-Cz. I know this might be hard to see so you might try drawing lines as you multiply each term, which is also a good way of making sure you don't leave anything out. So what I mean is you would draw a curved line from 2y^2 to y^2 and multiply them like you have done for the first step. Then draw a like from 2y^2 to 4y and multiply them, then 2y^2 to 5, then 3y to y^2, then 3y to 4y, and so on. This is also known as an extended version of FOIL (First, Outside, Inside, Last). If you haven't heard of it, it is a method for multiply to factors together, similar to what you have here. Hope this helps! Jaz EDIT: I'm not sure why all of the text is in a giant paragraph. I spaced it out so it would be easier to read but I guess I can't post like that, sorry!
Roger B. | Math and Science Tutor Focused on Deeper UnderstandingMath and Science Tutor Focused on Deeper...
5.0 5.0 (114 lesson ratings) (114)

Hi Maralyn,

Students often find this kind of problem hard because they have learned the FOIL (First, Outer, Inner, Last) method for simplifying problems like (x + 2)(x - 5) and don't know how to extend the FOIL method to problems like this one that have more than two terms in one or both of the expressions inside the parentheses.  But let's think for a moment about what the FOIL method is doing: it is multiplying each term from the first group by each term of the second group.  Well, that's exactly what you want to do for this problem: multiply each term of the first group by each term of the second group.  Of course, you also need to know how to multiply things like 2y2 by y2 which requires you to know that ym*yn = ym+n.  So, 2y2 * y2 = 2y4.

Hope this helps,