Question

NO LINKS!!! NOT MULTIPLE CHOICE!!!!!

1. Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies the given condition.

a. -5, 2, 1; f(3) = 32
f(x) = ___________

b. -3i, -3i, 2; f(1) = 30
f(x) = ___________


b.

Answer 1

`\textsf{a)} \quad f(x)=2x^3+4x^2-26x+20`

`\textsf{b)} \quad f(x)=-3x^3+6x^2-27x+54`

Explanation:

Factor Theorem

If f(x) is a polynomial, and f(c) = 0, then (x – c) is a factor of f(x).

Part a

Given information:

• Degree: 3
• Zeros: -5, 2, 1
• Condition: f(3) = 32

According to the factor theorem the polynomial in factored form is:

`f(x)=a(x-(-5))(x-2)(x-1)`

`f(x)=a(x+5)(x-2)(x-1)`

To find the value of the leading coefficient, a, apply the condition:

`\begin{aligned}f(3)=a(3+5)(3-2)(3-1)&=32\\a(8)(1)(2)&=32\\16a&=32\\a&=2\end{aligned}`

Therefore, the factored polynomial is:

`f(x)=2(x+5)(x-2)(x-1)`

Expand the polynomial:

`f(x)=2(x^2+3x-10)(x-1)`

`f(x)=2(x^3-x^2+3x^2-3x-10x+10)`

`f(x)=2x^3+4x^2-26x+20`

Part b

Given information:

• Degree: 3
• Zeros: -3i, 3i, 2
• Condition: f(1) = 30

According to the factor theorem the polynomial in factored form is:

`f(x)=a(x-(-3i))(x-3i)(x-2)`

`f(x)=a(x+3i)(x-3i)(x-2)`

To find the value of the leading coefficient, a, apply the condition:

`\begin{aligned}f(1)=a(1+3i)(1-3i)(1-2)&=30\\a(1-9i^2)(-1)&=30\\a(1-9(-1))(-1)&=30\\-10a&=30\\a&=-3\end{aligned}`

Therefore, the factored polynomial is:

`f(x)=-3(x+3i)(x-3i)(x-2)`

Expand the polynomial:

`f(x)=-3(x^2-9i^2)(x-2)`

`f(x)=-3(x^2+9)(x-2)`

`f(x)=-3(x^3-2x^2+9x-18)`

`f(x)=-3x^3+6x^2-27x+54`