Using the rational zeros theorem to find all zeros of a polynomial

Question

Using the rational zeros theorem to find all zeros of a polynomial  

The function below has at least one rational zero.  Use this fact to find all zeros of the function
       h(x)=7x^4-9x^3-41x^2+13x+6
if more than one zero, separate with commas.  Write exact values, not decimal approximations

Answer 1

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

 

 

 

 

 

 

 

 

Answer 2

The rational root theorem says that, in this case, the numerator of the root must be a factor of 6 and the denominator of the root must be a factor of 7.

 

Since you are told that the polynomial has at least 1 rational route, it must be a fraction with one of ±{6,1,3,2} in the numerator and one of ±{7,1} in the denominator.

 

Hoping that the problem can be solved without a lot of work, try letting the denominator = 1.  You will find that there is a solution for numerators of 3 and -2. Thus, two rational roots are 3 and -2, and their corresponding factors are (x-3) and (x+2).

 

(x-3)(x+2)=x^2-x-6.

 

Divide the polynomial by x^2-x-6

 

(7x^4-9x^3-41x^2+13x+6)/( x^2-x-6)=7x^2-2x-1

 

Solve 7x^2-2x-1=0 to obtain the remaining two solutions.

 

x = 1/7 (1 - 2 Sqrt[2])

 

x = 1/7 (1 + 2 Sqrt[2])