To find a polynomial function of degree 3 with real coefficients given the zeros, we start with the zeros: 3, 1, and 4. We can express the polynomial in factored form as:
f(x) = a(x – 3)(x – 1)(x – 4)
Here, a is a constant that we need to determine using the condition f(2) = 24.
First, we can substitute x = 2 into the polynomial:
f(2) = a(2 – 3)(2 – 1)(2 – 4)
This simplifies to:
f(2) = a(-1)(1)(-2) = 2a
Now, we set this equal to 24 as given:
2a = 24
To solve for a, we divide both sides by 2:
a = 12
Now that we have the value of a, we can write our polynomial function:
f(x) = 12(x – 3)(x – 1)(x – 4)
To express it in standard form, we can expand it:
f(x) = 12[(x – 3)(x – 1)(x – 4)]
First, we expand (x – 3)(x – 1):
(x – 3)(x – 1) = x^2 – 4x + 3
Next, we multiply this result by (x – 4):
(x^2 – 4x + 3)(x – 4) = x^3 – 4x^2 – 4x^2 + 16x + 3x – 12 = x^3 – 8x^2 + 19x – 12
Now we bring back the 12:
f(x) = 12(x^3 – 8x^2 + 19x – 12) = 12x^3 – 96x^2 + 228x – 144
Thus, the polynomial function we were looking for is:
f(x) = 12x^3 – 96x^2 + 228x – 144