anonymous
  • anonymous
An expression is shown below: 2x3y + 8xy − 4x2y − 16y Part A: Rewrite the expression so that the GCF is factored completely. (4 points) Part B: Rewrite the expression completely factored. Show the steps of your work. (6 points)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@pooja195
Jacob902
  • Jacob902
A) 2x3y + 8xy - 4x2y - 16y looks like we can take out a 2y so 2y(x3 + 4x - 2x2 - 8)
Jacob902
  • Jacob902
B) 2y(x3 + 4x - 2x2 - 8) group like this 2y (x3 - 2x2) + (4x-8) then factor x^2 and 4 2Y x2(x-2) +4(x-2) factor out x-2 (2y)(x-2)(x2+4) C) what do you think??

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anonymous
  • anonymous
there is no c @Jacob902
Jacob902
  • Jacob902
anyways cool awesome hope you understand
princeharryyy
  • princeharryyy
2x^2y(x-2) +8y(x-2) => (x-2)(2x^2y+8y) =>2y(x-2)(x^2+4)
anonymous
  • anonymous
is that for a or b? @princeharryyy
princeharryyy
  • princeharryyy
they are both.
anonymous
  • anonymous
so the same answer @princeharryyy
Jacob902
  • Jacob902
I assume "2x3y + 8xy - 4x2y - 16y" means 2x³y + 8xy - 4x²y - 16y. If not, ignore my answer. Part A: 2y(x³ + 4x - 2x² - 8) Part B: 2y(x(x² + 4) - 2(x² + 4)) = 2y(x - 2)(x² + 4)
princeharryyy
  • princeharryyy
See>> for A)) for the first part as we don't know what the values of x and y are, they are both variables the highest common factor can be any of the three 2y or (x-2) or (x^2+4) but as you know (x-2) is always less than (x^2+4) the GCF can only be 2y or (x^2+4) or (y or 2*(x^2+4))
princeharryyy
  • princeharryyy
So, the answer most probably cannot be determined, because the answer can be from any of those two.
princeharryyy
  • princeharryyy
for the second part >>> 2x^2y(x-2) +8y(x-2) => (x-2)(2x^2y+8y) =>2y(x-2)(x^2+4) was all >>> :) @mackenzieee I hope u r getting, what I am trying to say.
anonymous
  • anonymous
okay I see now I am just a little confused on if that is showing my work for part b
princeharryyy
  • princeharryyy
write three steps that's all >> the stuff that you need.
anonymous
  • anonymous
okay thank you
princeharryyy
  • princeharryyy
thanks @Jacob902
princeharryyy
  • princeharryyy
u r welcome @mackenzieee

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