Question

NEED HELP

Answer 1

Answer:

$\displaystyle \frac{1}{2}(x^3-2x^2-5x+6)$

Step-by-step explanation:

Polynomials

a)

The polynomial whose graph is shown is of third degree because it has three real roots. The roots of a polynomial are the values of x that make the expression equal to zero. We can see it happens three times in the graph provided. The roots or zeros are  

x=-2, x=1, x=3

b)

The factored form of a polynomial whose roots $x_1, x_2, x_3$ are known is

$a(x-x_1)(x-x_2)(x-x_3)$

We know the value of the roots, thus the polynomial is written as

$a(x+2)(x-1)(x-3)$

We need to find the value of a. We do that by replacing the value of x=0 and finding a that makes f(0)=3 (as seen in the graph). Thus

$a(0+2)(0-1)(0-3)=3$

$a(2)(-1)(-3)=3$

$a(6)=3$

$\displaystyle a=\frac{3}{6}$

$\displaystyle a=\frac{1}{2}$

Thus the factored form of the polynomial is

$\displaystyle \frac{1}{2}(x+2)(x-1)(x-3)$

c)

Let's multiply all the factors

$\displaystyle \frac{1}{2}(x+2)(x-1)(x-3)$

$\displaystyle \frac{1}{2}(x^2+2x-x-2)(x-3)$

$\displaystyle \frac{1}{2}(x^2+x-2)(x-3)$

$\displaystyle \frac{1}{2}(x^3-3x^2+x^2-3x-2x+6)$

$\boxed{ \displaystyle \frac{1}{2}(x^3-2x^2-5x+6)}$