- Factors of 6 are (leading coefficient and constant) : \(\pm {1}, \pm {2}, \pm {3}, \pm {6}\)
So all the possible roots \(\pm {1}, \pm{\dfrac{1}{2}}, \pm{\dfrac{1}{3}}, \pm{\dfrac{1}{6}}, \pm{\dfrac{2}{3}}, \pm {3}, \pm{6}\)
- 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: \(\pm{1}, \pm{\dfrac{1}{2}}, \pm{\dfrac{1}{4}}, \pm{\dfrac{3}{2}}, \pm{\dfrac{3}{4}},\pm{9}\,\pm{\dfrac{9}{2}}, \pm{\dfrac{9}{4}}\)
- 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 \(\pm { 1}, \pm {2}, \pm {4}, \pm{\dfrac{1}{2}}, \pm{\dfrac{1}{3}}, \pm{\dfrac{1}{6}},\pm{\dfrac{2}{3}},\pm{\dfrac{4}{3}}\)
- 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: \(3\),\(\dfrac{1}{2}\) and \(\dfrac{2}{3}\)