Find all the zeros of the function f(g) = g^4 + 2g^3 + 10g^2 + 18g + 9

Question

Find all the zeros of the function f(g) = g^4 + 2g^3 + 10g^2 + 18g + 9

Answer 1

This polynomial can be expressed in the factored form (g-a)(g-b)(g-c)(g-d) + 9 = 0. Thus, abcd must equal -9. From the rational roots test, the set of possible zeroes is:

 

{ -3,+3,-1,+1,-9,+9,sqrt(-9), -sqrt(-9), sqrt(-3), -sqrt(-3), sqrt(-1), -sqrt(-1), sqrt(+1), -sqrt(1), sqrt(+3), -sqrt(3), sqrt(+9), sqrt(-9) }. How did I know that the square roots were possible? Because of the highest  power (exponent) is even. Since this is a 4th level polynomial, 4th roots are also possible, but I want to keep solving the problem as simple as possible and would only try them if the roots are not all found. Notice that the set contains complex numbers and can be rewritten as

 

{ -1, +1, -3  +3, -9, +9, -3i, +3i, -sqrt(3)i, +sqrt(3)I, -I, I }

 

Because all coefficients are positive, no root can be a positive integer because the sum of the first four terms must be a negative number (-9 to be exact). That reduces the set to

 

{ -1, -3,  -9, -3i, +3i, -sqrt(3)i, +sqrt(3)i, -i, +i }

 

Based on this, synthetic division can be used to find the roots, thusly:

 

-1 ||  1   2  10  18  -9

            -1  -1   -9    9

        1    1   9    9    0

 

So -1 is a factor. Based on synthetic division, -3 and -9 are not factors. Synthetic division is not limited to integers.

 

3i ||  1   1             9     9

           3 I       -9+3i   -9

        1   1+3i        3i    0

 

So 3i is a factor.

     

-3i ||  1     1 + 3i     3i

                     -3i    -3i

         1      1    1      0

 

So -3i is a factor. There are no other factors. The polynomial can therefore be shown in factored form as:

 

(g + 1)(g – 3i)(gn + 3i)(g – d) + 9 = 0.

 

To match the 0th degree term, this means that (1)(-3i)(+3i)(d) = -9. The only way that can be true is if g = -1 is double-poled, i.e. d = +1.

 

So, the zeroes of the polynomial are -1, -1, 3i, -3i, and the factored form is

 

(g + 1)^2 * (g – 3i) * (g + 3i) + 9 = 0

Answer 2

f(g) = g^4 + 2g^3 + 10g^2 + 18g + 9 can be rewritten as f(g)= (g+1)^2*(g^2+9). to find the zeros set f(g)=0 and you get that f(g)=0 when g=-1 (real solution), g=3i, and g=-3i (complex solutions). 

Answer 3

The rational roots theorem says that the possible rational roots of this quartic are +1, -1, + 3, -3, +9, -9.

Trying all six gets just one hit  g = -1

 

This means that the quartic is a product of (g+1) and a cubic polynomial.  The cubic polynomial can be found by either synthetic division or long division.   The result is   g³ + g² + 9 g + 9

 

This cubic can be analyzed by a method called factoring by grouping.    The first two terms and the last two have a common factor of g + 1.   So the original quartic can be written as

 

(g + 1) (g + 1) x quadratic_function.   The quadratic can be found by a number of method (including long division).  It is g² + 9

Putting this all together gives

 

f(g) = (g + 1) ( g+1 ) (g² + 9 )

 

Thus there is just one root at g = -1 which is a double root.