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0001007831 Group Title

60x^2-68x+8

  • 2 years ago
  • 2 years ago

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    4(15x^2-17x+2) 4(15x^2-15x-2x+2) 4(15x(x-1)-2(x-1)) 4(15x-2)(x-1))

    • 2 years ago
  2. 0001007831 Group Title
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    thank you do you mind if u do 7x^3-14x^2-3x+6

    • 2 years ago
  3. lindseywashere1010 Group Title
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    7x^2(x-2) - 3(x-2) (7x^2 - 3)(x-2)

    • 2 years ago
  4. 0001007831 Group Title
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    thanks again but if u really do not mind can u please do 30x^2+95xy+50y^2 and2x^2+xy-3y^2 and x^3-26x^2+25 thanks a million u r a lifesaver i have a teacher who can not teach and i do not learn anything. Again thank you

    • 2 years ago
  5. Hero Group Title
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    I figured that maybe you would have understood the method by now

    • 2 years ago
  6. 0001007831 Group Title
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    I only understand a few but some are really diffucult that I have no clue how to start

    • 2 years ago
  7. Hero Group Title
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    The first step is to always see if you can find something common to each term. If so, factor it out. Make sure you look at the coefficients and the variables

    • 2 years ago
  8. Hero Group Title
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    Once you do that, what's left inside the parentheses may be in a form where you can factor it.

    • 2 years ago
  9. 0001007831 Group Title
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    soo can u show me step by step how to solve 30x^2+95xy+50y^2

    • 2 years ago
  10. Hero Group Title
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    The first step is to find the highest factor common to each term...(coefficients and variables). In this case, only the coefficients have a common factor. And that factor is 5. So we factor it out: 5(6x^2 + 19xy + 10y^2) Now we're left with something simpler inside the parentheses. The part in the parentheses above is all that is left that we need to factor. So, The next step is to find a number that multiplies to get 60 but add to get 19. By doing this, it will allow us to split the middle term into two factors so that we can factor by grouping. By trial and error, those factors are 4 and 15. 4 + 15 = 19 4*15 = 60 So now, we re-write the middle term below like so: 5(6x^2+4xy + 15xy + 10y^2) Now, we have to factor by grouping. To do that, we simply factor out what's common to the first two terms inside the parentheses: 6x^2 and 4xy, then what's common to the last two terms inside the parentheses: 15xy and 10y^2 For the first two terms, 2x is common For the last two terms, 5y is common: 5(2x(3x+ 2y)+5y(3x+2y)) Now, once again, paying attention only to what's inside the parentheses, we see that 3x+2y is common to two inner terms inside the parentheses. Notice that 2x+5y are outside the (3x+2y). So when we factor out (3x+2y) we get this: 5((3x+2y)(2x+5y)) And that's the final answer. Sorry it took so long to explain.

    • 2 years ago
  11. 0001007831 Group Title
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    thats ok that really helped me and i think i am starting to understand thanks again

    • 2 years ago
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