Fine a third degree polynomial function f(x) with real coefficients that has 4 and 2i are zeros and such that f(-1) =-50
4 21 -2i
(x-4)(x-2i)(x+2i)
(x-4)(x2+4)
x3-16+4x-4x2
(x3-4x2+4x-16)
-50=a(-1-4-4-16)
-50=.25
a=2
Not understanding, please help
I assume you get this far ok:
...
(x-4)(x^2+4)
x^3-4x^2+4x-16
evaluate that at x = -1, and you get
-1-4-4-16 = -25
That means that if
f(x) = a(x^3-4x^2+4x-16)
then f(-1) = -25a
But, we want f(-1) = -50
so, -25a = -50
a = 2
f(x) = 2(x^3-4x^2+4x-16)
Sure! Allow me to explain it to you in a more understandable way.
Given that the zeros of the polynomial are 4 and 2i, we know that the polynomial has three factors: (x - 4), (x - 2i), and (x + 2i). Since these factors correspond to the zeros, we can rewrite the polynomial as follows:
f(x) = (x - 4)(x - 2i)(x + 2i)
Now let's simplify the expression:
f(x) = (x - 4)(x^2 + 4)
Expanding this further, we get:
f(x) = x^3 + 4x - 4x^2 - 16
Combining like terms, we find:
f(x) = x^3 - 4x^2 + 4x - 16
To find the value of the constant 'a', we use the given information that f(-1) = -50:
f(-1) = (-1)^3 - 4(-1)^2 + 4(-1) - 16
Simplifying this expression gives:
-1 + 4 - 4 - 16 = a(-1 - 4 - 4 - 16)
-17 = a(-25)
To solve for 'a', we divide both sides of the equation by -25:
a = -17 / -25
Simplifying this further, we get:
a = 17/25
Therefore, the third-degree polynomial function with real coefficients that satisfies the given conditions is:
f(x) = (17/25)(x^3 - 4x^2 + 4x - 16)
I hope that clears things up! If you have any more questions, feel free to ask.
To find the third degree polynomial function with real coefficients that has the given zeros and satisfies the given condition, follow these steps:
Step 1: Start by assuming the polynomial function is of the form f(x) = a(x - r1)(x - r2)(x - r3), where r1 = 4 and r2 = 2i are the given zeros.
Step 2: Since the coefficients are real, complex zeros occur in conjugate pairs. Therefore, the third zero must be the conjugate of 2i, which is -2i.
Step 3: Substitute the given zeros into the equation: f(x) = a(x - 4)(x - 2i)(x + 2i).
Step 4: Simplify the equation: f(x) = a(x - 4)(x^2 - (2i)^2).
Step 5: Simplify further: f(x) = a(x - 4)(x^2 + 4).
Step 6: Expand the equation: f(x) = a(x^3 + 4x - 4x^2 - 16).
Step 7: Rearrange the terms: f(x) = ax^3 - 4ax^2 + 4ax - 16a.
Step 8: Given that f(-1) = -50, substitute x = -1 and f(x) = -50 into the equation.
Step 9: Solve for the value of 'a': -50 = a(-1)^3 - 4a(-1)^2 + 4a(-1) - 16a.
Step 10: Simplify: -50 = -a - 4a + 4a - 16a.
Step 11: Combine like terms: -50 = -17a.
Step 12: Solve for 'a': a = -50 / -17.
Step 13: Simplify: a ≈ 2.941.
Therefore, the third degree polynomial function with real coefficients that satisfies the given conditions is f(x) ≈ 2.941x^3 - 11.764x^2 + 11.764x - 47.059.
To find the polynomial function that satisfies the given conditions, we can start by using the zeros and the fact that the polynomial has real coefficients.
Given that the zeros are 4 and 2i (where i is the imaginary unit), we know that the polynomial must have these factors: (x - 4) and (x - 2i) and (x + 2i).
Multiplying these factors together, we have:
(x - 4)(x - 2i)(x + 2i)
Next, we can simplify these factors.
(x - 4)(x - 2i)(x + 2i)
(x - 4)(x^2 - (2i)^2)
(x - 4)(x^2 - 4i^2)
(x - 4)(x^2 + 4)
Expanding the last expression, we get:
x^3 + 4x - 4x^2 - 16
Rearranging the terms, we have the final third-degree polynomial function:
f(x) = x^3 - 4x^2 + 4x - 16
Now, to verify the value of f(-1) = -50, we substitute x = -1 into the function:
f(-1) = (-1)^3 - 4(-1)^2 + 4(-1) - 16
f(-1) = -1 + 4 - 4 - 16
f(-1) = -17
This means that f(-1) is equal to -17, not -50. Therefore, there seems to be an error in the calculation. It's important to double-check the calculations to ensure accurate results.