POLQ 3| Lesson 2| Practice (Rational Roots)

1).  Factors of 6 are ( leading coefficient and constant) : \(\pm {1}, \pm {2}, \pm {3}, \pm {6}\)

So all the possible roots \(\pm {1}, \pm{\frac{1}{2}}, \pm{\frac{1}{3}}, \pm{\frac{1}{6}}, \pm{\frac{2}{3}}, \pm {3}, \pm{6}\)

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

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

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

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

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

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

4.  Find the roots for problem 1

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:  \(3\),\(\frac{1}{2}\) and \(\frac{2}{3}\)

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