Question

Construct a polynomial function with the started properties. Third-degree, with zeros of −3 , − 1 , and 2 , and passes through the point (3,5)

Answer 1

Step 1: Convert zeros to factors of the polynomial.

If "k" is a zero, then (x-k) is a factor of the polynomial.

The allows us to convert zeros of -3, -1, and 2 into the factors (x+3)(x+1)(x-2).

So right now we have y= (x+3)(x+1)(x-2), but there's a missing piece, the leading coefficient of "a", so really we currently have

y = a · (x+3)(x+1)(x-2)

This is where the point (3,5) comes in. We need to substitute x=3 and y=5 into the equation above to find a.

5 = a · (3+3)(3+1)(3-2)

5 = a · (6)(4)(1)

5 = a · 24

5/24 = a

That's the full function:

f(x) = 5/24 (x+3)(x+1)(x-2)

Answer 2

`f(x)=(5)/(24)x^3+(5)/(12)x^2-(25)/(24)x-(5)/(4)`

Explanation:

The zeros of a polynomial f(x) are the values of x which satisfy f(x) = 0.

If the polynomial function is degree 3 with 3 zeros, in factored form it can be written as f(x) = k(x - r₁)(x - r₂)(x - r₃) where r are the zeros and k ≠ 0.

Substitute the given zeros -3, -1 and 2 into the formula:

`\implies f(x)=k(x-(-3))(x-(-1))(x-2)`

`\implies f(x)=k(x+3)(x+1)(x-2)`

To find the value of k, substitute the given point (3, 5) into the function:

`\begin{aligned}\implies f(3)&=5\\k(3+3)(3+1)(3-2)&=5\\k(6)(4)(1)&=5\\24k&=5\\k&=(5)/(24)\end{aligned}`

Therefore, the factored form of the polynomial function is:

`\implies f(x)=(5)/(24)(x+3)(x+1)(x-2)`

To write in standard form, expand and distribute:

`\implies f(x)=(5)/(24)(x^2+4x+3)(x-2)`

`\implies f(x)=(5)/(24)(x^3+2x^2-5x-6)`

`\implies f(x)=(5)/(24)x^3+(10)/(24)x^2-(25)/(24)x-(30)/(24)`

`\implies f(x)=(5)/(24)x^3+(5)/(12)x^2-(25)/(24)x-(5)/(4)`