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anonymous

  • 5 years ago

25xto the 3rd degreey to 2degree +125xto 2 degreeyto5degree FACTOR COMPLETELY

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  1. anonymous
    • 5 years ago
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    \[(25x^3)^2+(125x^2)^5\]?

  2. anonymous
    • 5 years ago
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    HOW DO YOU ENTER EXPONENTS ON THE KEYBOARD

  3. anonymous
    • 5 years ago
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    i think you have a symol palette where it says \[\sum\]equation

  4. anonymous
    • 5 years ago
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    i am writing latex to do it

  5. anonymous
    • 5 years ago
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    i would start by writing \[(5^2x^3)^2+(5^3x^2)^5\]

  6. anonymous
    • 5 years ago
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    if this is the problem.

  7. anonymous
    • 5 years ago
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    if it is not let me know and i will not continue with this method

  8. anonymous
    • 5 years ago
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    NO WHEN I PUT TO WHATEVER DEGREE THAT IS THE EXPONENT

  9. anonymous
    • 5 years ago
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    25XTO 3 DEGREE Y TO 2 DEGREE + 125X TO 2 DEGREE Y TO 5 DEGREE

  10. anonymous
    • 5 years ago
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    \[25(x^3)^2+125(x^2)^5\]?

  11. anonymous
    • 5 years ago
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    YOU LEFT OUT HE Y'S AND THEIR DEGREES AND NO BRACKETS OR PARENTHESES

  12. anonymous
    • 5 years ago
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    oho

  13. anonymous
    • 5 years ago
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    \[25x^3y^2+125x^2y^5\]

  14. anonymous
    • 5 years ago
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    YES

  15. anonymous
    • 5 years ago
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    ok each has a common factor of \[25x^2y^2\] is that clear?

  16. anonymous
    • 5 years ago
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    so you can factor it out and write \[25x^2y^2(x+5y^3\]

  17. anonymous
    • 5 years ago
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    NO, I NEED STEP BY STEP

  18. anonymous
    • 5 years ago
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    ok. you have two numbers 25 and 125

  19. anonymous
    • 5 years ago
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    YES

  20. anonymous
    • 5 years ago
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    their greatest common factor is 25 because \[25=25\times 1\] and \[125=25\times 5\] so that is going to come out of the parenthese

  21. anonymous
    • 5 years ago
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    just looking at that part we can say that \[25+125=25(1+5)\]

  22. anonymous
    • 5 years ago
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    THERE IS NO PARENTHESES

  23. anonymous
    • 5 years ago
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    no there are not. but "factoring" means to write as a produce. that is our job, to put the parenthese in

  24. anonymous
    • 5 years ago
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    OK

  25. anonymous
    • 5 years ago
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    so now for the variables: first term has \[x^3\] second term has \[x^2\]

  26. anonymous
    • 5 years ago
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    their greatest common factor is \[x^2\] because \[x^2=x^2\times 1\] and \[x^3=x^2\times x\]

  27. anonymous
    • 5 years ago
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    I AM SORRY BUT THIS ISN'T HELPING

  28. anonymous
    • 5 years ago
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    so we are going to "factor out" a 25 and an \[x^2\]

  29. anonymous
    • 5 years ago
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    i have a better idea.

  30. anonymous
    • 5 years ago
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    look at my answer which is \[25x^2y^2(x+5y^3)\] multiply out using the distributive law and see if you get what you started with.

  31. anonymous
    • 5 years ago
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    maybe then it will be clear where the \[25x^2y^2\] came from, and why we pulled it out front of the parentheses

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