Question

Using factor theorem,show that g(x) is a factor of p(x),when
$g(x) = x + \sqrt{2}$
$p(x) = 2 \sqrt{2} x {}^{2} + 5x + \sqrt{2}$

Answer 1

if so then
$x = - \sqrt{2}$
satisfies p(x)
$= 2 \sqrt{2} ( - { \sqrt{2}) }^{2} + 5( - \sqrt{2} ) + \sqrt{2} \\ = 2 \sqrt{2} \times 2 - 5 \sqrt{2} + \sqrt{2} \\ = 4 \sqrt{2} + \sqrt{2} - 5 \sqrt{2} \\ = 5 \sqrt{2} - 5 \sqrt{2} \\ = 0$
hence g(x) is factor of p(x)