Question

Find all the zeros of the equation. -x^3-3x^2=6x+4

Answer 1

Answer:

The zeros of given expression $-x^3-3x^2=6x+4$ is -1, $-1+i\sqrt{3}$ and $-1-i\sqrt{3}$

Step-by-step explanation:

Given expresssion is $-x^3-3x^2=6x+4$

To find zeros of given expression we have to equate the expression to zero.

ie., $-x^3-3^2-6x-4=0$

$-(x^3+3x^2+6x+4)=0$

$x^3+3x^2+6x+4=0$

By using synthetic division

-1  |  1    3    6    4

    |  0   -1   -2   -4

    |________________

       1    2   4    0

Therefore (x+1) is a  zero

Now the quadratic equation is $x^2+2x+4=0$

For quadratic equation $ax^2+bx+c=0$  we have

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

Here a=1 ,b=2 and c=4 now substitute the values

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

$x=\frac{-2\pm \sqrt{4-16}}{2}$

$x=\frac{-2\pm \sqrt{-12}}{2}$

$x=\frac{-2\pm \sqrt{12i^2}}{2}$ where $i^2=-1$

$x=\frac{-2\pm i \sqrt{4\times 3}}{2}$

$x=\frac{-2\pm i \sqrt{4}\sqrt{3}}{2}$

$x=\frac{-2\pm 2i\sqrt{3}}{2}$

$x=2\times\frac{(-1\pm i\sqrt{3})}{2}$

$x=-1\pm i\sqrt{3}$

Therefore $x=-1+i\sqrt{3}$  and $x=-1-i\sqrt{3}$

Therefore the zeros are -1, $-1+i\sqrt{3}$ and $-1-i\sqrt{3}$