toxicsugar22
  • toxicsugar22
Find the indicated term in the product. (3x+x^3-7x^5)(1+4x-3x^2) for x^3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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toxicsugar22
  • toxicsugar22
can u help me
toxicsugar22
  • toxicsugar22
can someone help
toxicsugar22
  • toxicsugar22
blexting can u help me

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More answers

anonymous
  • anonymous
Sure
anonymous
  • anonymous
There is a long way and a short way. the long way is to multiply this out.. you would distribute 3 times.
anonymous
  • anonymous
Or. you could look at how can I get x^3 by multiplying terms. If you are not that good at math we should do it the long way.
anonymous
  • anonymous
(3x + x^3 - 7x^5) (1) = 3x + x^3 - 7x^5 (3x + x^3 - 7x^5)(4x) = 12 x^2 + 4x^4 - 28x^6 (3x + x^3 - 7x^5)(-3x^2) = -9x^3 - 3x^5 + 21x^7 Now we only look at the x^3 terms. I see a x^3 and -9x^3 so if you add those we get -8x^3
toxicsugar22
  • toxicsugar22
ok
toxicsugar22
  • toxicsugar22
that is the answer
anonymous
  • anonymous
Yes
toxicsugar22
  • toxicsugar22
thanks and can u help me with one more
anonymous
  • anonymous
Sure
toxicsugar22
  • toxicsugar22
ok
toxicsugar22
  • toxicsugar22
it bassicly the same thing
toxicsugar22
  • toxicsugar22
is that okay
anonymous
  • anonymous
Yes
toxicsugar22
  • toxicsugar22
Find the indicated term in each product (4+2x-x^2)(2-3x+2x^2) for x^2
anonymous
  • anonymous
(4 + 2x - x^2)(2) = 8 + 4x - 2x^2 (4 + 2x - x^2)(-3x) = -12x - 6x^2 + 3x^3 (4 + 2x - x^2)(2x^2) = 8x^2 + 4x^3 - 2x^4 do you understand how I get these?
toxicsugar22
  • toxicsugar22
yes
anonymous
  • anonymous
OK.. now look at all the terms on the right. typically we would add all of these up to get the answer, but they only want the x^2 term
anonymous
  • anonymous
So I see... -2x^2 and -6x^2 and 8x^2
anonymous
  • anonymous
So add those up. -2x^2 - 6x^2 + 8x^2 = 0x^2 so actually you have no x^2 terms in the answer. So I guess I would say 0x^2
toxicsugar22
  • toxicsugar22
wow thanks i got it
toxicsugar22
  • toxicsugar22
i gave u a medal
anonymous
  • anonymous
Thanks... good luck

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