math.answers.com/math-and-arithmetic/How_do_you_simplify_x2-9_over_x2-6x_plus_9

Preview meta tags from the math.answers.com website.

Linked Hostnames

8

Thumbnail

Search Engine Appearance

Google

https://math.answers.com/math-and-arithmetic/How_do_you_simplify_x2-9_over_x2-6x_plus_9

How do you simplify x2-9 over x2-6x plus 9? - Answers

First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)



Bing

How do you simplify x2-9 over x2-6x plus 9? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_simplify_x2-9_over_x2-6x_plus_9

First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)



DuckDuckGo

https://math.answers.com/math-and-arithmetic/How_do_you_simplify_x2-9_over_x2-6x_plus_9

How do you simplify x2-9 over x2-6x plus 9? - Answers

First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)

  • General Meta Tags

    22
    • title
      How do you simplify x2-9 over x2-6x plus 9? - Answers
    • charset
      utf-8
    • Content-Type
      text/html; charset=utf-8
    • viewport
      minimum-scale=1, initial-scale=1, width=device-width, shrink-to-fit=no
    • X-UA-Compatible
      IE=edge,chrome=1

  • Open Graph Meta Tags

    7
    • og:image
      https://st.answers.com/html_test_assets/Answers_Blue.jpeg
    • og:image:width
      900
    • og:image:height
      900
    • og:site_name
      Answers
    • og:description
      First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)

  • Twitter Meta Tags

    1
    • twitter:card
      summary_large_image

  • Link Tags

    16
    • alternate
      https://www.answers.com/feed.rss
    • apple-touch-icon
      /icons/180x180.png
    • canonical
      https://math.answers.com/math-and-arithmetic/How_do_you_simplify_x2-9_over_x2-6x_plus_9
    • icon
      /favicon.svg
    • icon
      /icons/16x16.png

Links

58