101k views
3 votes
Third-degree, with zeros of -3,-1, and 2 and passes through the point (3,6)

Third-degree, with zeros of -3,-1, and 2 and passes through the point (3,6)-example-1
User Melvio
by
8.2k points

1 Answer

6 votes

Since the polynomial must have zeroes at x=-3, x=-1, x=2, then, we can write it as a combination of the factors (x+3), (x+1), (x-2):


p(x)=k(x+3)(x+1)(x-2)

The constant k will help us to adjust the value of the polynomial when x=3:


\begin{gathered} p(3)=k(3+3)(3+1)(3-2) \\ =k(6)(4)(1) \\ =24k \end{gathered}

Since p(3) must be equal to 6, then:


\begin{gathered} 24k=6 \\ \Rightarrow k=(6)/(24) \\ \Rightarrow k=(1)/(4) \end{gathered}

Therefore, the following polynomial function has zeroes at -3, -1 and 2, and passes through the point (3,6):


p(x)=(1)/(4)(x+3)(x+1)(x-2)

User Afrosteve
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories