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steve816

  • one year ago

Is it possible to factor this without the use of a calculator?

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  1. steve816
    • one year ago
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    \[x^3+x^2-5x+3\]

  2. zepdrix
    • one year ago
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    You can apply your `Rational Root Theorem` to look for "possible roots". The leading coefficient is a 1 on the x^3. Factors of 1 include 1 and -1. The constant on the end is a 3. Factors of 3 include \(\large\rm \color{orangered}{1, 3}\) and \(\large\rm\color{orangered}{-1, -3}\). Rational Root Theorem tells us to to take the combination of the constant factors and divide by the leading coefficient factors, individually. These will give us our possible roots which are rational. Here is the list for this specific polynomial:\[\large\rm \pm\frac{\color{orangered}{1}}{1},\pm\frac{\color{orangered}{-1}}{1},\pm\frac{\color{orangered}{3}}{1},\pm\frac{\color{orangered}{-3}}{1}\]I skipped any that would repeat.

  3. zepdrix
    • one year ago
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    So IF this polynomial has a rational root, it has to be \(\large\rm 1,-1,3\text{ or }-3\).

  4. zepdrix
    • one year ago
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    Let's check x=1, nice simple process.\[\large\rm x^3+x^2-5x+3\qquad\to\qquad (1)^3+(1)^2-5(1)+3\]This results in 0. Good! We've found one of our roots / zeroes / solutions of this polynomial. You can apply polynomial long division or synthetic division to factor it down to a quadratic.

  5. steve816
    • one year ago
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    Wow, great explanation! Thanks for your time and effort :)

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