# For the function g(x) = -\frac{1}{2}(2x-1)(x+3)^3(x+1) a. Find the degree of the polynomial...

## Question:

For the function {eq}g(x) = -\frac{1}{2}(2x-1)(x+3)^3(x+1) {/eq}

a. Find the degree of the polynomial function

b. Find the leading term

c. Find the x- and y- intercepts.

## The Degree and Leading Coefficient of a Polynomial

A polynomial is a function which can be written in the form {eq}f(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \ldots + a_2x^2 +a_1x + a_0 {/eq} where {eq}n {/eq} is a positive integer, {eq}a_n\neq 0 {/eq} and {eq}a_0, a_1, \ldots a_n {/eq} are constants. The degree of a polynomial is the highest exponent of all of its terms. So the polynomial {eq}f(x) = 4x^8 + 23x^2 + 13x +2 {/eq} has degree 8. The leading coefficient of a polynomial is the constant multiplied onto the x with the highest power. In the polynomial mentioned before, the leading coefficient is 4.

## Answer and Explanation:

For {eq}g(x) = -\frac{1}{2}(2x-1)(x+3)^3(x+1) {/eq}

a) To find the degree of the polynomial, we could multiply out all of the terms, but this is not necessary. Simply imagine multiplying out all of the terms to find the highest exponent. The highest exponent will come from multiplying {eq}-\frac{1}{2}(2x) {/eq}, {eq}x^3 {/eq}, and {eq}x {/eq}, resulting in {eq}-x^5 {/eq}, so the degree of the polynomial is 5.

b) The leading term is the term with the highest power. From the work from part a, we know this is {eq}-x^5 {/eq}

c) The x-intercepts occur when y=0 and the y-intercept occurs when x=0. For the x-intercepts, we have to solve

{eq}0 = -\frac{1}{2}(2x-1)(x+3)^3(x+1) {/eq}

Setting each factor equal to zero, we have

{eq}0=2x-1\\ 1=2x\\ \frac{1}{2}=x {/eq}

and {eq}0=(x+3)^3\\ 0=x+3\\ -3=x {/eq}

and {eq}0=x+1\\ -1=x {/eq}

So the x-intercepts are {eq}\left(\frac{1}{2}, 0\right) , (-3, 0) , (-1,0) {/eq}

To find the y-intercept, substitute in x=0

{eq}g(0) = -\frac{1}{2}(2(0)-1)(0+3)^3(0+1)\\ g(0) = -\frac{1}{2}(-1)(27)(1)\\ g(0) = \frac{27}{2} {/eq}

So the y-intercept is {eq}\left(0, \frac{27}{2}\right) {/eq}