Exercise 48 Page 78

There is only one sign change for the coefficients of f(- x). Once again, according to Descartes' Rule of Signs, the function has one negative real zero. Therefore, there are two options for the types of zeros of f(x). Remember, we know that there are four zeros in total.

Step 3

Let's now determine the real zeros by making a table of values.

We see above that a zero occurs at x=- 1. Let's use synthetic division to depress the polynomial.

The above are the coefficients of the depressed polynomial. Moreover, since - 1 is a root of f(x)=0, (x+1) is a factor of f(x). f(x)=x^4-6x^3+7x^2+6x-8 ⇕ f(x)=(x+1) ( x^3-7x^2+14x-8 ) From the table, we know that another zero occurs at x=1. Thus, (x-1) is a factor of f(x). We can use synthetic division again to rewrite the depressed polynomial. If you want to see how to use synthetic division with the depressed polynomial, please see the end of this exercise. f(x)=(x+1) ( x^3-7x^2+14x-8 ) ⇕ f(x)=(x+1) (x-1) ( x^2-6x+8 ) Now the depressed is a quadratic polynomial, where a= 1, b= - 6, and c= 8. This means we can use the Quadratic Formula to find its zeros.

Let's now find the third and fourth zeros by splitting the final fraction into the positive and negative cases.

Since x=4 and x=2 are zeros of the depressed polynomial, we can rewrite them as (x-4)(x-2). f(x)=(x+1) (x-1) ( x^2-6x+8 ) ⇕ f(x)=(x+1)(x-2)(x-4)(x-2) The four zeros are - 1, 1, 4, and 2.

Graphing the Polynomial

We will use the zeros and the points obtained in the table to graph the polynomial function. Consider also that this is an even-degree polynomial with a positive leading coefficient. This tells us about the end behavior of the function. &f(x) → + ∞ as x → - ∞ &f(x) → + ∞ as x → + ∞ Let's draw the function.