For the polynomial For the polynomial, -2 is a zero. $h(x)= x^3+8x^2+14x+4$. Express $h(x)$ as a product of linear factors. 
Can someone please explain and help me solve? 
 A: Apply the Euclidean division of $x^3+8x^2+14x+4$ and $x+2$ and you will get:
$$x^3+8x^2+14x+4=(x+2) \cdot (x^2+6x+2)$$
Then find the roots of $x^2+6x+2=0$
A: Hint $\ $ You can infer the sum and product of the other 2 roots because you know the sum and product of all 3 roots (from the coefficients, by Vieta).
A: Since its a cubic equation, you are looking at $$(x+A).(x+B).(x+C)=x^3+8x^2+14x+4$$
Obviously, you have to multiply and work hard towards solving this and getting the answer.
Alternatively, what you can do is represent the same thing as
$$x^3+8x^2+12x+2x+4=0$$
Or, $$x(x^2+8x+12)+2(x+2)$$
Or, $$x(x+6)(x+2)+2(x+2)$$
Or, $$(x+2)(x(x+6)+2)$$
Or, $$(x+2)(x^2+6x+2)$$
Solving the second equation for x will give $$x= -3 +\sqrt{7}$$
and $$x=-3-\sqrt{7}$$
and $$x=-2$$ 
A: Hint: $a$ is a root of the polynomial $f(x)$ if and only if the polynomial $x-a$ divides $f(x)$. So if you divide $h(x)$ by $x+2$ you get a polynomial of degree $2$. Do you know how to find the roots of a polynomial of degree $2$?
A: Okay, the first thing to do is polynomial division (or synthetic division, whichever you prefer).
Since -2 is a 0, we know that $(x+2)$ is a factor of $h(x)$. We then divide $h(x)$ by $(x+2)$
Dividing: $$\frac{x^3 + 8x^2 + 14x + 4}{x+2} = x^2 + 6x + 2$$
So $h(x)$ becomes: $$(x+2)*(x^2+6x+2)$$
Now you must use the Quadratic formula to find the root of $x^2 + 6x +2$
$$x = \frac{-6 +/- \sqrt{28}}{2}$$
$$x = -3 +/- \sqrt{7}$$
So $h(x)$ is $$(x+3+\sqrt{7})(x+3-\sqrt{7})(x+2)$$
