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

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I* need to know how to find all rational zeros of polynomials.
asked Mar 10, 2014 in ALGEBRA 2 by mathgirl Apprentice

1 Answer

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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  = -1 + √(3)i  and 

x  = -1 - √(3)i.

answered Mar 10, 2014 by ashokavf Scholar
edited Mar 10, 2014 by ashokavf

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