# 008A Sample Final A, Question 9

Question: a) List all the possible rational zeros of the function $f(x)=x^{4}+5x^{3}-27x^{2}+31x-10$ b) Find all the zeros, that is, solve f(x) = 0
Foundations:
1) What does the Rational Zeros Theorem say about possible zeros?
2) How do you check if a possible zero is actually a zero?
3) How do you find the rest of the zeros?
1) The possible divisors can be found by finding the factors of -10, in a list, and the factors of 1, in a second list. Then write down all the fractions with numerators from the first list and denominators from the second list.
2) Use synthetic division, or plug a possible zero into the function. If you get 0, you have found a zero.
3) After your reduce the polynomial with synthetic division, try and find another zero from the list you made in part a). Once you reach a degree 2 polynomial you can finish the problem with the quadratic formula.

Solution:

Step 1:
Start by factoring -10, and 1. Then the Rational Zeros Theorem gives us that the possible rational zeros are $\pm 1,\pm 2,\pm 5,$ and $\pm 10$ .
Step 2:
Start testing zeros with 1 and -1 since they require the least arithmetic. You will also find that 1 is a zero. Applying synthetic division you can reduce the polynomial to $(x-1)(x^{3}+6x^{2}-21x+10)$ .
Step 3:
Now we just need to find the zeros of $x^{3}+6x^{2}-21x+10$ . Since we are not down to a quadratic polynomial we have to find another zero from the list of rational zeros we found in step 1. You will find 2 is another root, and the polynomial can further be reduced to $(x-1)(x-2)(x^{2}+8x-5)$ Step 4:
Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are ${\frac {-8\pm {\sqrt {64+20}}}{2}}={\frac {-8\pm {\sqrt {4\cdot 21}}}{2}}=-4\pm {\sqrt {21}}$ . Thus the zeros of $x^{4}+5x^{3}-27x^{2}+31x-10$ are $1,2,$ and $-4\pm {\sqrt {21}}$ The roots are $x=1,2,$ and $-4\pm {\sqrt {21}}$ 