Please help and explain how to do it

Question

you have to find the number of positive and negative real zeros, including complex. Use the equation below

f(x)=x^5-3x^4-5x^3+5x^2-6x+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.

e. Write the polynomial in factor form.

f. What are the complex zeros?

Answer 1

Part B

The Rational Root Theorem states that the possible rational zeroes of a polynomial are by taking the factors of the constant 8 and dividing it by the factors of the leading coefficient 1. Therefore, the possible rational zeroes are ±(8,4,2,1)/±(1) = ±8,±4,±2,±1

Part C

Descartes' Rules of Signs tells us that the number of sign changes from left to right in f(x) tell how many possible positive real zeroes there are, and then the number of sign changes in f(-x) for negative real zeroes.

There are 4 sign changes in f(x), so there are either 4, 2, or 0 possible positive real zeroes

Since f(-x) = (-x)⁵-3(-x)⁴-5(-x)³+5(-x)²-6(-x)+8 = -x⁵-3x⁴+5x³+5x²+6x+8, and there is only 1 sign change in f(-x), then there is 1 negative real zero.

If there are 4 positive real zeroes and 1 negative real zero, there are no complex zeroes

If there are 2 positive real zeroes and 1 negative real zero, there are 2 complex zeroes

If there are 0 positive real zeroes and 1 negative real zero, there are 4 complex zeroes

Part D

When f(x) is graphed, there are 3 rational zeroes, which are -2, 1, and 4. You don't necessarily have to use synthetic division to check, but if you plug in f(-2), f(1), and f(4), you should get 0 for each.

Part E+F

Since x=-2 is a zero, then (x+2) is a factor of f(x)

Since x=1 is a zero, then (x-1) is a factor of f(x)

Since x=4 is a zero, then (x-4) is a factor of f(x)

(x⁵-3x⁴-5x³+5x²-6x+8)/(x+2) = x⁴-5x³+5x²-5x+4

(x⁴-5x³+5x²-5x+4)/(x-1) = x³-4x²+x-4

(x³-4x²+x-4)/(x-4) = x²+1

x²+1 = 0

x² = 1

x = ±i

Therefore, the polynomial in factored form is (x+2)(x-1)(x-4)(x+i)(x-i), and the complex zeroes are x=±i

Hope this helped!