Find an nth-degree polynomial function with real coefficients satisfying the given conditions.
n=4;
i and 3 i are zero;
f(-2)=65
f(x)=
An expression using x as the variable. Simplify your answer.
all irrational and complex zeros come in conjugal pairs,
so if i is a zero , so is -i
if 3i is a zero , so is -3i
so we have f(x) = a(x-i)(x+i)(x-3i)(x+3i)
f(x) = a(x^2 + 1)(x^2 + 9)
also f(-2) = 65
a(4+1)((4+9) = 65
65a = 65
a = 1
f(x) = (x^2 + 1)(x^2 + 9)
or
f(x) = x^4 + 10x^2 + 9
check:
http://www.wolframalpha.com/input/?i=solve+x%5E4+%2B+10x%5E2+%2B+9%3D0
Sure, here's an nth-degree polynomial function that satisfies the given conditions:
f(x) = (x - i)(x + i)(x - 3i)(x + 3i)
Since 3i and -3i are also zeros, we can include them in the factors. Now, let's simplify this expression:
f(x) = (x^2 + 1)(x^2 + 9)
Applying the FOIL method to multiply these quadratics:
f(x) = x^4 + 10x^2 + 9
Therefore, the nth-degree polynomial function with real coefficients satisfying the given conditions is:
f(x) = x^4 + 10x^2 + 9
To find an nth-degree polynomial function f(x) with real coefficients that satisfies the given conditions, we know that the complex conjugate i and 3i are both zeros of the polynomial. This means that (x - i) and (x + i) are factors, as well as (x - 3i) and (x + 3i) since complex zeros always appear in conjugate pairs.
So, we have (x - i)(x + i) and (x - 3i)(x + 3i) as factors. We can multiply these factors to find the polynomial:
(x - i)(x + i) = x^2 + i^2 = x^2 - 1
(x - 3i)(x + 3i) = x^2 + (3i)^2 = x^2 - 9i^2 = x^2 + 9
Multiplying these factors together gives us:
(x^2 - 1)(x^2 + 9) = x^4 + 9x^2 - x^2 - 9 = x^4 + 8x^2 - 9
Now we need to find the constant term, which we can do using the given condition f(-2) = 65.
Substitute x = -2 into the polynomial and set it equal to 65:
(-2)^4 + 8(-2)^2 - 9 = 16 + 32 - 9 = 39.
We need to adjust the constant term to be 65 instead of 39. To do this, we add 26 to our polynomial:
f(x) = x^4 + 8x^2 - 9 + 26
Simplifying the polynomial function, we get:
f(x) = x^4 + 8x^2 + 17
Therefore, the nth-degree polynomial function that satisfies the given conditions is f(x) = x^4 + 8x^2 + 17.
To find the nth-degree polynomial function, we can use the fact that if i and 3i are zeros, then their conjugates -i and -3i are also zeros. Therefore, we can express the polynomial function as:
f(x) = a(x - i)(x + i)(x - 3i)(x + 3i)
Since all the zeros are complex, the coefficients of the polynomial function will be real.
Let's first simplify the equation:
(x - i)(x + i) = (x^2 - i^2) = (x^2 + 1)
(x - 3i)(x + 3i) = (x^2 - (3i)^2) = (x^2 + 9)
Now, we can substitute these simplifications into the equation:
f(x) = a(x^2 + 1)(x^2 + 9)
To find the value of 'a', we can use the given condition:
f(-2) = 65
Substituting -2 for x:
65 = a((-2)^2 + 1)((-2)^2 + 9)
65 = a(4 + 1)(4 + 9)
65 = a(5)(13)
65 = 65a
Dividing both sides by 65, we get:
a = 1
Now we can substitute 'a' back into the equation:
f(x) = (x^2 + 1)(x^2 + 9)
That is the nth-degree polynomial function satisfying all the given conditions.