Question

Is x-2 a factor of 3x cube-7x square+10x_ 16

Answer 1

hope this answer will help you.

Answer 2

Answer:

$(3x - \frac{1 + \iota\sqrt{95}}{2}),\ (x - \frac{1 - \iota\sqrt{95}}{6})$

Step-by-step explanation:

$3x^3 - 7x^2 + 10x - 16 \\ \\ = 3x^3 - 6x^2 - x^2 + 2x + 8x - 16 \\ \\ = 3x^2(x - 2) - x(x - 2) + 8(x - 2) \\ \\ = (3x^2 - x + 8)(x - 2) \\ \\ = (3x^2 - \frac{1 - \sqrt{-95}}{2}x - \frac{1 + \sqrt{-95}}{2}x + 8)(x - 2) \\ \\ = (3x(x - \frac{1 - \sqrt{-95}}{6}) - \frac{1 + \sqrt{-95}}{2}(x - \frac{1 - \sqrt{-95}}{6}))(x - 2) \\ \\ = (3x - \frac{1 + \sqrt{-95}}{2})(x - \frac{1 - \sqrt{-95}}{6})(x - 2) \\ \\ = (3x - \frac{1 + \iota\sqrt{95}}{2})(x - \frac{1 - \iota\sqrt{95}}{6})(x - 2)\ \ \ \ \ \ \ \ \ \ [\iota = \sqrt{-1}]$

$\therefore\ (3x - \frac{1 + \iota\sqrt{95}}{2}),\ (x - \frac{1 - \iota\sqrt{95}}{6})\ are the other factors. \\ \\ \\ \therefore\ x = 2\ \ \ ; \ \ \ x = \frac{1 + \iota\sqrt{95}}{6}\ \ \ ; \ \ \ x = \frac{1 - \iota\sqrt{95}}{6}$