@jim_thompson5910 I multiply the stuff in the first parenthesis by the stuff in the second parenthesis, right? http://www.sketchtoy.com/34803580
9 multiplies, then add "like terms"
9?!
yes. 9! what do you get for the first 3 : x^2 times each of the terms on the right side ?
Ohh, wait, I see why.
how did you get the red equation? the way to do it is start with x^2, and multiply x^2 times 2x^2 x +6 and write down the answer for each of those then multiply 4x times each of 2x^2 x +6 then multiply -3 times each of 2x^2 x +6
do you know how to do \[ x^2 \cdot 2x^2 \]? remember this is the same as \[ x\cdot x \cdot 2 \cdot x \cdot x \] and you can change the order of the multiplies to \[ 2 \cdot x\cdot x \cdot x \cdot x \] and use the "short-hand" x^4 instead of x*x*x*x \[ 2 x^4 \]
Give me some background information on what you're doing. It appears we're dealing with multiplication of trinomials.
The worksheet is actually called Multiplication of Polynomials.
Poly, whateva. Okay, so in order to multiply a trinomial we want to multiply every time by on another.
Then you just have to add like terms
So the red equation I have down already.. would I multiply that by 4x now?
Huh, the 4x in the original equation? You want to multiple every term...
No, the red equation in this: http://www.sketchtoy.com/34804492 is the product of multiplying x^2 by the original equation.
Then you multiply by -3
Whoa, where did that second part come from?
I multiplied 4x by everyting in the second equation.
Oh, okay! So then I multiply both the red and blue by -3?
No. You multiply everything by -3 on the 2nd parenthesis.
Are you aware of what we're even doing? :/
Yes, I'm just really really bad at it > . <
If I had (x + y + z) (x + y + z) I'd multiply everything by x x^2 + xy + xz Then I'd do the same for every other term yx + y^2 + yz And now time for z zx zy z^2 Does that make sense?
yes, i'll just wing it i guess.
and now add all the like terms?
This is actually really easy I just think you're making it more difficult than it actually is.. http://www.sketchtoy.com/34809753 I second, I'll draw something up for you
This is too confusing. I'm just going to ask my teacher. Thanks for the help though.
No no. I'm already drawing something. yOu will wait!
The first term in that initial multiplication is wrong: x^2 * 2x^2 is not 3x^4!
See what I did here? http://www.sketchtoy.com/34810550 This is literally one of the easiest things you can do in school :/
Here's the way to think about it: Remember the distributive property of multiplication? \[a(b+c) = ab + ac\] We just apply that several times: \[a(b+c+d) = ab + ac + ad\] \[(a+b+c)(d +e + f) = a(d+e+f) + b(d+e+f) + c(d+e+f) \]\[= ad + ae + af + b(d + e + f) + c(d + e + f)\]\[= ad+ae+af + bd + be + bf + c(d + e + f)\]\[= ad + ae + af + bd + be + bf + cd + ce +cf\]
So your problem: \[(x^2+4x-3)(2x^2+x+6) = x^2(2x^2+x+6) + 4x(2x^2+x+6) -3(2x^2+x+6)\]\[=2x^4+x^3+6x^2 + 8x^3+4x^2+24x -6x^2-3x-18\] Collect like terms and you're done!
Sorry if part of it went off the right side, I hope you get the idea...
@Compassionate That's what I've been doing, I was only one number off. And you know that very last number is a 6, right?
I thought it was b. Your handwriting is scary ;-; Yea, do what you did for the first one to every term in the first set of parentheses.
Sorry, homeschooling does horrible things to handwriting. > . < And okay.
correct. Now the same for -3
Last term in the red should be \(6x^2\) not \(6^2\)
Oops, thanks for catching that @whpalmer4
Now add like terms
I don't know how I would add the last three terms because they aren't like :/
is it 12x^2
warning, you lost the last part of the sum in that sketch...
I fixed it.
You're too quick for me :-)
She just gave up. Ha ha :P
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