anonymous
  • anonymous
Use the factor Theorem to factor the polynomial completely.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Given that 2 is a solution; 2x^3+x^2-13x+6=0
anonymous
  • anonymous
How would I go about solving this?
Nnesha
  • Nnesha
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

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Nnesha
  • Nnesha
like leading coefficient is 2 and factors 2 and 1 you will write it as \[\frac{ constant }{ leading ~coefficinet }=\frac{ ? }{ \pm 2, \pm 1 }\]
Nnesha
  • Nnesha
gtg sorry :(
anonymous
  • anonymous
I'm still confused. The example problem uses synthetic division in the beginning...
anonymous
  • anonymous
triciaal
  • triciaal
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
triciaal
  • triciaal
you already have the polynomial factored what do you need help with?
anonymous
  • anonymous
The screenshot is an example problem, I was just confused as to what the question was asking.
triciaal
  • triciaal
|dw:1441855710559:dw|
triciaal
  • triciaal
|dw:1441855967549:dw|
triciaal
  • triciaal
factor completely 2 x^3 +x^2-13 x + 6 = (x-2)(2 x-1)(x+3)
anonymous
  • anonymous
I see!
triciaal
  • triciaal
great
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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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