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anonymous

  • one year ago

whats the formula for factoring polynomial

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  1. triciaal
    • one year ago
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    you basically regroup and take out common terms then use the common factor for each group essentially more common terms do you have a specific example you want to work through?

  2. triciaal
    • one year ago
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    the remainder theorem says in my words if (x-a) is a factor of f(x) then when x = a the value of f(a) = 0

  3. triciaal
    • one year ago
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    if you divide p by q and there is no remainder then q is a factor of p

  4. triciaal
    • one year ago
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    |dw:1438058403700:dw|

  5. anonymous
    • one year ago
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    you try to solve the system of elementary symmetric polynomials that make the coefficients; for a quadratic, notice: $$(x-a)(x-b)=x^2-(a+b)x+ab$$ so in order to factor a quadratic of the form \(x^2+px+q\) we try to solve for \(a,b\) given \(p,q\): $$a+b=-p\\ab=q$$

  6. anonymous
    • one year ago
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    for example, if we have \(x^2+9x+14\), we want to find \(a,b\) such that: $$a+b=-9\\ab=14$$ so we could notice that \(a=-2,b=-7\) works giving \(x^2+9x+14=(x-(-2))(x-(-7))=(x+2)(x+7)\)

  7. anonymous
    • one year ago
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    normally you don't write it out, though, and instead just ask yourself mentally "what factors do I know of \(q\) that add up to \(p\)?", since: $$(x+a)(x+b)=x^2+(a+b)x+ab$$ so in \(x^2+9x+14\) we think "well, 2 and 7 make 14 and add to 9, so..."

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