Answer
a) $= (2x -1)(x-3)$
b) $= (2cos\theta -1)(cos\theta - 3)$
Work Step by Step
a) $2x^{2} -7x + 3$
$= 2x^{2} - 6x - 1x + 3$
$= 2x(x - 3) - 1(x - 3)$
$= (2x -1)(x-3)$
b) $2(cos\theta)^{2} -7cos\theta+3$
Let $x = cos\theta$ to get
$2x^{2} - 7x + 3$ (Exact same equation as in Section a))
$=(2x-1)(x-3)$.
Substitute $cos\theta$ into $x$ to get
$ (2cos\theta -1)(cos\theta - 3)$