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anonymous

  • one year ago

Express the rational expression in lowest terms: 3x+4/15x^2+26x+8.

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  1. Jhannybean
    • one year ago
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    by "lowest terms" you want to factor this?

  2. anonymous
    • one year ago
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    I think so, heres the question again with the choices:

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  3. Jhannybean
    • one year ago
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    Oh, alright. so you want to factor \(\sf 15x^2+26x+8\)

  4. anonymous
    • one year ago
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    \[\frac{4 x^2}{15}+29 x+8 \]

  5. Jhannybean
    • one year ago
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    i'd say your best choice would be to leave it as it is.,

  6. anonymous
    • one year ago
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    I"m not sure, but I just wanna know how to solve this

  7. Nnesha
    • one year ago
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    do you know how to factor the quadratic equations ???

  8. anonymous
    • one year ago
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    no please teach me

  9. Nnesha
    • one year ago
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    alright there are like 3 ways to factor the quadratic equation but the easy one is AC method you have to find two numbers when you multiply them you should get product of AC and when you add or subtract them you should get the middle term B \[\huge\rm Ax^2+Bx+C\] A=leading coefficient B=middle term C=constant term

  10. Nnesha
    • one year ago
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    |dw:1442180776707:dw| multiply A C

  11. Jhannybean
    • one year ago
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    But @Nnesha the factored terms will not cancel out with the numerator..

  12. Nnesha
    • one year ago
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    yes one of the factor will cancel out with the numerator.

  13. Nnesha
    • one year ago
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    A=15 B=26 C=8 multiply 15 times 8= 120 now find two number when you multiply them u should get 120 and when u add or subtract them you should get 26

  14. Nnesha
    • one year ago
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    what are the two numbers whose sum is 26 and product is 120 ???

  15. anonymous
    • one year ago
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    Is it 6 and 20?

  16. Nnesha
    • one year ago
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    yes right now we have to apply group method |dw:1442182184861:dw| make a group of two terms now find take out the common factor from first red box

  17. Nnesha
    • one year ago
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    15x^2+20x what is common in these two terms ?

  18. anonymous
    • one year ago
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    I"m not sure?

  19. Nnesha
    • one year ago
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    what are the factors of 15 and 20 ?

  20. Nnesha
    • one year ago
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    nvm 5 times 3 =15 5 times 4 = 20 so 5,3 are factors of 15 and 5,4 are factors of 20 and x^2 can be written as x times x so \[\huge\rm (\color{ReD}{5 · 3 ·x ·x} +\color{blue}{ 5 · 4 ·x})\] now can you tell me what's common in red and blue terms ?

  21. anonymous
    • one year ago
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    15 x squared is equal from the red, and 20x is blue?

  22. Nnesha
    • one year ago
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    yes it's 15x^2 +20x but i factored 15 , and 20 so you can find the common factor nvm so \[\huge\rm (\color{ReD}{5} · 3 ·x ·\color{ReD}{x} +\color{red}{ 5} · 4 ·\color{ReD}{x})\] now as you can see 5 and x are common in both terms right ? so take them out \[\huge\rm \color{ReD}{5x}(3x+4)\] understood ?

  23. Nnesha
    • one year ago
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    15x^2 = first term 20x = 2nd term just in case if you're not familiar with it :=)

  24. Jhannybean
    • one year ago
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    \[15x^2+26x+8 \implies x^2+26x+120\]\[=(x+6)(x+20)\]\[=\left(x+\frac{6}{15}\right)\left(x+\frac{20}{15}\right)\]\[=\left(x+\frac{\cancel{6}2}{\cancel{15}5}\right)\left(x+\frac{\cancel{20}4}{\cancel{15}3}\right)\]\[\sf \text{cross multiply} : (5x+2)(3x+4)\]\[=\frac{3x+4}{(5x+2)(3x+4)}\] So what would eliminating \(\sf 3x+4\) from the top and bottom give you?

  25. Nnesha
    • one year ago
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