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The function below has at least one rational zero. Use this fact to find ALL zeros of the function.

g(x)=5x^3+x^2-20^x-4

If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

4 zeros

Step-by-step explanation:

If you want to use the Rational Zeros Theorem, as instructed, you need to use synthetic division to find zeros until you get a quadratic remainder.

P: ±1, ±2, ±3, ±6  (all prime factors of constant term)

Q: ±1, ±7              (all prime factors of the leading coefficient)

P/Q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7  (all possible values of P/Q)

Now, start testing your values of P/Q in your polynomial:

f(x)=7x4-9x3-41x2+13x+6

You can tell f(1) and f(-1) are not zeros since they're not = 0Now try f(2) and f(-2):

f(2)=7(16)-9(8)-41(4)+13(2)+6

       112-72-164+26+6 ≠ 0

f(-2)=7(16)-9(-8)-41(4)+13(-2)+6

       112+72-164-26+6 = 0  OK!! There is a zero at x=-2

This means (x+2) is a factor of the polynomial.

Now, do synthetic division to find the polynomial that results from

(7x4-9x3-41x2+13x+6)÷(x+2):

-2⊥ 7   -9   -41    13     6

         -14   46   -10    -6          

      7  -23   5       3     0     The remainder is 0, as expected

The quotient is a polynomial of degree 3:

7x3-23x2+5x+3

Now, continue testing the P/Q values with this new polynomial.  Try f(3):

f(3)=7(27)-23(9)+5(3)+3

      189-207+15+3 = 0  OK!!  we found another zero at x=3

Now, another synthetic division:

3⊥ 7   -23    5     3

          21   -6   -3_

    7    -2    -1    0    

The quotient is a quadratic polynomial:

7x2-2x-1  This is not factorable, you need to apply the quadratic formula to find the 3rd and 4th zeros:

x= (1±2√2)÷7  

The polynomial has 4 zeros at x=-2, 3, (1±2√2)÷7

45

step-by-step explanation: