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anonymous

  • one year ago

Use the factor Theorem to factor the polynomial completely.

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  1. anonymous
    • one year ago
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    Given that 2 is a solution; 2x^3+x^2-13x+6=0

  2. anonymous
    • one year ago
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    How would I go about solving this?

  3. Nnesha
    • one year ago
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    i guess it's p over q method \[\huge\rm \frac{ p }{ q }= \frac{ constant }{ leading ~coefficient }\] find factors of leading coefficient and factors of constant term

  4. Nnesha
    • one year ago
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    like leading coefficient is 2 and factors 2 and 1 you will write it as \[\frac{ constant }{ leading ~coefficinet }=\frac{ ? }{ \pm 2, \pm 1 }\]

  5. Nnesha
    • one year ago
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    gtg sorry :(

  6. anonymous
    • one year ago
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    I'm still confused. The example problem uses synthetic division in the beginning...

  7. anonymous
    • one year ago
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  8. triciaal
    • one year ago
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    given that 2 is a solution then (x-2) is a factor. (x-2) is a factor means when you divide the polynomial by the factor there is no remainder. Divide by (x-2) then factor the quadratic

  9. triciaal
    • one year ago
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    you already have the polynomial factored what do you need help with?

  10. anonymous
    • one year ago
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    The screenshot is an example problem, I was just confused as to what the question was asking.

  11. triciaal
    • one year ago
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    |dw:1441855710559:dw|

  12. triciaal
    • one year ago
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    |dw:1441855967549:dw|

  13. triciaal
    • one year ago
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    factor completely 2 x^3 +x^2-13 x + 6 = (x-2)(2 x-1)(x+3)

  14. anonymous
    • one year ago
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    I see!

  15. triciaal
    • one year ago
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    great

  16. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Thanasi99 Given that 2 is a solution; 2x^3+x^2-13x+6=0 \(\color{blue}{\text{End of Quote}}\) ahh i was just looking at the equation didn't read the statement D:D

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