Use the equation below to answer a series of questions with a complete explanation (this is SAT prep)

Question

f(x)=x⁵-3x⁴-5x³+5x²-6x+8

b. Identify all possible rational zeros.

c. How many possible positive real zeros are there? How many possible negative real zeros? How many possible complex zeros?

d. Graph the polynomial to approximate the zeros. What are the rational zeros? Use synthetic division to verify these are correct. (I graphed it myself but can you explain how to use synthetic division here)

e. Write the polynomial in factor form.

f. What are the complex zeros?

Answer 1

a. Given the Rational Root Theorem, all possible rational zeros of a polynomial are some factor of the unit term (in this case, 8) divided by some factor of the leading term's coefficient (in this case, 1). Therefore, all possible rational zeroes would be -8,-4,-2,-1,1,2,4,8.

b.

Descartes' Rule of Signs says that the maximum number of positive real zeros is, at most, the number of sign changes of successive terms of f(x). f(x) changes sign four times here (between x^5 and -3x^4, between -5x^3 and 5x^2, between 5x^2 and -6x, and between -6x and 8). Therefore, there are at most 4 real positive roots.

Descartes' Rule of Signs says that the maximum number of negative real zeros, is, at most, equal to the number of sign changes of f(-x).

f(-x) will be equal to (-x)^5-3(-x)^4-5(-x)^3+5(-x)^2-6(-x)+8, which is equal to

-x^5-3x^4+5x^3+5x^2+6x+8. There is only one sign change here (between -3x^4 and 5x^3), so there is at most 1 negative real root.

All complex roots MUST come in pairs (ie a-bi has to be accompanied by a+bi). This equation has a degree of 5. 4 is the largest even number that is less than 5, therefore the equation has at most 4 complex roots.

c. By graphing out our equation, we find that our roots are at x=-2, x=1, and x=4.

d. To use synthetic division, we first use x=1 as our first zero.

1 | 1 -3 -5 5 -6 8

| 1 -2 -7 -2 8

-------------------------------------------

1 -2 -7 -2 -8 0

meaning that we have (x-1)(x^4-2x^3-7x^2-2x-8)

We then then use synthetic division for this resulting polynomial, using x=-2 as our next zero, using the same method from above.

-2 | 1 -2 -7 -2 -8

| -2 8 -2 8

-----------------------------------------------------------------

-2 -4 1 -4 0

Thus, our equation becomes (x-1)(x+2)(x^3-4x^2+x-4)

We finally use synthetic division on our polynomial, with x=4 as our zero

This gives us the resulting equation (x-1)(x-4)(x+2)(x^2+1)

We cannot factor out (x^2+1) into real factors, so this is our final answer for part d.

e. Our complex zeros will be what we get when we plug x^2+1 into the Quadratic Formula.

Our imaginary roots are therefore i and -i