Question

Please help with this Math Question...

Answer 1

Answer: A

Step-by-step explanation:

f(x) = x³ + 4x² + 7x + 6

possible rational roots are ±{1, 2, 3, 6}

Try x = -2

-2 |   1   4   7   6

   |   ↓  -2  -4  -6

       1    2    3   0 ← remainder is 0 so x = -2 is a root  ⇒  (x + 2) = 0

The factored polynomial x² + 2x + 3 = 0 is not factorable so use the quadratic formula to find the roots.

a=1, b=2, c=3

$x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$

$x=\dfrac{-(2)\pm\sqrt{(2)^2-4(1)(3)} }{2(1)}$

$x=\dfrac{-2\pm\sqrt{4-12} }{2}$

$x=\dfrac{-2\pm\sqrt{-8} }{2}$

$x=\dfrac{-2\pm2i\sqrt{2}}{2}$

$x = -1 \pm i\sqrt{2}$

$x = -1 + i\sqrt{2}$       $x = -1 - i\sqrt{2}$

$x - (-1 + i\sqrt{2})=0$       $x - (-1 - i\sqrt{2})=0$

The factors are:

$(x - 2)[x - (-1 + i\sqrt{2})][x - (-1 - i\sqrt{2})]$