Question

5: Find the equation of a polynomial with given zeros

Answer 1

Given the zeros:

• -1 with a multiplicity of 2

,

• -3

,

• -5

Let's find the equation of the polynomial.

Using the given zeros, we have the following:

x = -1, x = -1, x = -3, x = -5

Now, rewrite each zero as a factor:

`\begin{gathered} x=-1 \\ Add\text{ 1 to both sides:} \\ x+1=-1+1 \\ x+1=0 \\ \\ x=-1 \\ x+1=0 \\ \\ x=-3 \\ \text{ Add 3 to both sides:} \\ x+3=-3+3 \\ x+3=0 \\ \\ \\ x=-5 \\ \text{ Add 5 to both sides:} \\ x-5+5=-5+5 \\ x+5=0 \end{gathered}`

Now, we have the equation:

`f(x)=(x+1)(x+1)(x+3)(x+5)`

Let's expand the equation usng the FOIL method and distributive property:

`\begin{gathered} f(x)=((x+1)(x+1))(x+3)(x+5) \\ \\ f(x)=(x(x+1)+1(x+1))(x+3)(x+5) \\ \\ f(x)=(x^2+x+x+1)(x+3)(x+5) \end{gathered}`

Solving further:

`\begin{gathered} f(x)=(x^2+2x+1)(x+3)(x+5) \\ \\ f(x)=(x(x^2+2x+1)+3(x^2+2x+1))(x+5) \\ \\ f(x)=(x^3+2x^2+x+3x^2+6x+3)(x+5) \\ \\ \end{gathered}`

Combine like terms:

`\begin{gathered} f(x)=(x^3+2x^2+3x^2+x+6x+3)(x+5) \\ \\ f(x)=(x^3+5x^2+7x+3)(x+5) \end{gathered}`

Now, apply FOIL method and distributive property once more:

`\begin{gathered} f(x)=x(x^3+5x^2+7x+3)+5(x^3+5x^2+7x+3) \\ \\ f(x)=x^4+5x^3+7x^2+3x+5x^3+25x^2+35x+15 \\ \\ \text{ Combine like terms:} \\ f(x)=x^4+5x^3+5x^3+7x^2+25x^2+3x+35x+15 \\ \\ f(x)=x^4+10x^3+32x^2+38x+15 \end{gathered}`

Therefore, the equation of a polynomial with the given zeros is:

`f(x)=x^4+10x^3+32x^2+38x+15`

• ANSWER: C

`f(x)=x^(4)+10x^(3)+32x^(2)+38x+15`