Question

A degree 3 polynomial that has zeros at x = -1, 2; f(-2) > 0 and f(0) < 0Find the missing degree and the possible answer.

Answer 1

ANSWER and EXPLANATION

The polynomial has a degree of 3. This means that it has 3 roots.

Let the roots of the polynomial be a, b, and c such that:

`(x-a)(x-b)(x-c)=0`

We are given two roots already (-1, 2). This implies that:

`(x+1)(x-2)(x-c)=0`

Expand the equation:

`\begin{gathered} (x^2-x-2)(x-c)=0_{} \\ \Rightarrow x^3-cx^2-x^2+cx-2x+2c=0 \\ x^3-(c+1)x^2+(c-2)x+2c=0^{} \end{gathered}`

We are given the following conditions:

`\begin{gathered} f(-2)>0 \\ f(0)<0 \end{gathered}`

Applying the first condition:

`\begin{gathered} (-2)^3-(c+1)(-2)^2+(c-2)(-2)+2c>0 \\ -8-4c-4-2c+4+2c>0 \\ \Rightarrow-8-4c>0 \\ \Rightarrow4c<-8 \\ c<-2 \end{gathered}`

Applying the second condition:

`\begin{gathered} (0)^3-(c+1)(0)+(c-2)(0)+2c<0 \\ \Rightarrow2c<0 \\ c<0 \end{gathered}`

Therefore, we have that:

`\begin{gathered} c<0;c<-2 \\ \Rightarrow c<-2 \end{gathered}`

Therefore, the third root can take any values that are less than -2.