To find a polynomial function of degree 3 with real coefficients given the zeros at 3, 1, and 4, we can start by expressing the polynomial in its factored form:
Let the polynomial be represented as:
f(x) = a(x – 3)(x – 1)(x – 4)
Here, a is a constant that we need to determine. The factors correspond to the zeros of the polynomial, meaning that when you substitute these values into the polynomial, it will yield zero.
Next, we will use the condition given that f(2) = 24 to find the value of a. We will substitute x = 2 into the polynomial:
f(2) = a(2 – 3)(2 – 1)(2 – 4)
Now, let’s calculate this step-by-step:
- Calculate (2 – 3) = -1
- Calculate (2 – 1) = 1
- Calculate (2 – 4) = -2
Substituting these back into the function:
f(2) = a(-1)(1)(-2) = 2a
Since we know f(2) = 24, we set up the equation:
2a = 24
Now solve for a:
a = 24 / 2 = 12
Now that we have the value of a, we can write the final polynomial:
f(x) = 12(x – 3)(x – 1)(x – 4)
We could expand this polynomial to get it in standard form:
- First, multiply (x – 3)(x – 1) = x^2 – 4x + 3
- Next, multiply this result by (x – 4):
- (x^2 – 4x + 3)(x – 4) = x^3 – 4x^2 – 4x^2 + 16x + 3x – 12
- Combine like terms: x^3 – 8x^2 + 19x – 12
- Finally, multiply by 12:
f(x) = 12(x^3 – 8x^2 + 19x – 12) = 12x^3 – 96x^2 + 228x – 144
So, the polynomial function we are looking for is:
f(x) = 12x^3 – 96x^2 + 228x – 144
And this function meets both conditions: it has real coefficients, zeros at 3, 1, and 4, and satisfies f(2) = 24.