POLQ 3 | Lesson 2 | Practice (Rational Roots) Solutions

  1. Factors of 6 are (leading coefficient and constant) : \(\pm {1}, \pm {2}, \pm {3}, \pm {6}\)
  2. So all the possible roots \(\pm {1}, \pm{\dfrac{1}{2}}, \pm{\dfrac{1}{3}}, \pm{\dfrac{1}{6}}, \pm{\dfrac{2}{3}}, \pm {3}, \pm{6}\)

  1. Factors of 9 are:  \(\pm {1}, \pm{3}, \pm{9}\)
  2. Factors of 4 are: \(\pm { 1}, \pm {2}, \pm {4}\)

    So the possible roots are:  \(\pm{1}, \pm{\dfrac{1}{2}}, \pm{\dfrac{1}{4}}, \pm{\dfrac{3}{2}}, \pm{\dfrac{3}{4}},\pm{9}\,\pm{\dfrac{9}{2}}, \pm{\dfrac{9}{4}}\)

  1. Factors of 4 are: \(\pm { 1}, \pm {2}, \pm {4}\)
  2. Factors of 6 are:  \(\pm {1}, \pm {2}, \pm {3}, \pm {6}\)

    So the possible roots are \(\pm { 1}, \pm {2}, \pm {4}, \pm{\dfrac{1}{2}}, \pm{\dfrac{1}{3}}, \pm{\dfrac{1}{6}},\pm{\dfrac{2}{3}},\pm{\dfrac{4}{3}}\)

  1. Find the roots for problem 1
  2. Try 3 as a root and it works as shown below because it has a remainder of 0.

    Factoring the quadratic:  \(6x^2-7x+2 = (2x-1)(3x-2)\)

    Therefore the roots of the equation are:  \(3\),\(\dfrac{1}{2}\) and \(\dfrac{2}{3}\)


Return to Practice (Rational Roots)

Go onto Lesson 3: Polynomials features: multiplicity, degree, end behavior