Answer: 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
