Once you have found one zero, use Quadratic Formula to find the other two zeros.
the first zero is 2
the equation is X^3 + 6x^2+8x - 48

Question

Once you have found one zero, use Quadratic Formula to find the other two zeros.
the first zero is 2
the equation is X^3 + 6x^2+8x - 48

anonymous · Answer

$$\begin{align*}x^3+6x^2+8x-48&=x^2(x+6)+8(x-6)\&=x^2(x-2+8)+8(x-2-4)\&=x^2(x-2)+8(x-2)+8x^2-32\&=(x-2)(x^2+8)+8(x^2-4)\&=(x-2)(x^2+8)+4(x-2)(x+2)\&=(x-2)(x^2+8+4(x+2))\&=(x-2)(x^2+8+4x+8)\&=(x-2)(x^2+4x+16)\end{align*}$$Divide through by $x-2$, which is the factor that made $x=2$ a zero:$$\frac{x^3+6x^2+8x-48}{x-2}=x^2+4x+16$$Now, we use the quadratic formula:$$\begin{align*}x&=\frac{-4\pm\sqrt{16-4(16)}}2\&=-2\pm\sqrt{1-4}\&=-2\pm3i\end{align*}$$

anonymous · Answer

Ooops, I divided incorrectly... hmm