what are the real zeros to the polynomial: P(x)=x^3+3x^2+6x+4

Question

I* need to know how to find all rational zeros of polynomials.

Answer 1

P (x) = x^3 + 3x^2 + 6x + 4

The Rational Zero Theorem can be used for finding the some possible zeros to test.

The leading coefficient is 1  and constant is 4.

The possible values of p are   ± 1, ±2 , ±4.

Make a table for the synthetic division and test possible  zeros.

p/q	1	3	6	4
1	1	4	10	14
-1	1	2	4	0
2	1	5	16	36
-2	1	1	4	-4
4	1	7	34	140
-4	1	-1	10	-36

Since P (-1) =  0,  x  = -1 is a zero. The depressed polynomial is x^2 + 2x + 4 = 0. 

Since the depressed polynomial of this zero, x^2 + 2x + 4, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation.

x  = [-b ± √(b^2 - 4ac)] / 2a

a  = 1, b  = 2, c  = 4.

x  = [-2 ± √(4-16)] / 2

x  = [-2 ± √-12] / 2

x  = [-2 ± √4(-3)] / 2

x  = [-2 ± 2√(3)i] / 2

x  = 2(-1 ± √(3)i) / 2

x  = -1 ± √(3)i

The function has one real zero at x = -1 and two imaginary zeros at x  = -1 + √(3)i  and 

x  = -1 - √(3)i.