Question

Using factor theorem, show that g(x) is a factor of p(x), when p(x) = 2x(to the power 4) + x cube - 8x sq. -x+6, g(x) = 2x-3

Answer 1

from g(x), put the value of x in p(x)
$2x - 3 = 0 \\ 2x = 3 \\ x = \frac{3}{2} \\ p(x) =$
$p(x) = 2 {x}^{4} + {x}^{3} - 8 {x}^{2} - x + 6 \\ p( \frac{3}{2}) = 2( { \frac{3}{2} )}^{4} + ({ \frac{3}{2} )}^{3} - 8( { \frac{3}{2} )}^{2} - \frac{3}{2} + 6 \\ = 2 \times \frac{81}{16} + \frac{27}{8} - 8 \times \frac{9}{4} - \frac{3}{2} + 6 \\ = \frac{81}{8} + \frac{27}{8} -1 8 - \frac{3}{2} + 6 \\ = \frac{108}{8} - \frac{3}{2} - 12 \\ = \frac{108 - 12 - 96}{8} \\ = \frac{108 - 108}{8} \\ = \frac{0}{8} \\ = 0$
so remainder comes zero,that means g(x) is
factor of p(x).