Could you please check my answers?

Find an nth degree polynomial function with real coefficients satisfying the given conditions.

1. n=3; 3 and i are zeros; f(2)=20
-I got: f(x)=-4^3+12x^2-4x+12

3.n=3;4 and i zeros;f(-3)=60
-I got:f(x)=6x^3+24x^2+6x+24

To determine whether your answers are correct, let's check them step-by-step.

1. For n = 3 with zeros 3 and i, the polynomial function can be written as:

f(x) = (x - 3)(x - i)(x + i)

To simplify this expression, we can use the fact that "i" is a complex zero, so its conjugate "-i" must also be a zero. Hence, the expression becomes:

f(x) = (x - 3)(x - i)(x + i)(x + (-i))

Expanding this equation:

f(x) = (x^2 - 3x - xi + 3i)(x^2 - xi + 3x + 3i)
= (x^2 - 3x - xi + 3i)(x^2 + xi + 3x - 3i)
= (x^2 - 3x + 3)(x^2 + 3x + 3)

To match the given condition f(2) = 20, we can substitute x = 2 into the expression:

f(2) = (2^2 - 3(2) + 3)(2^2 + 3(2) + 3)
= (4 - 6 + 3)(4 + 6 + 3)
= (1)(13)
= 13

Hence, your answer of f(x) = -4^3 + 12x^2 - 4x + 12 is incorrect.

2. For n = 3 with zeros 4 and i, the polynomial function can be written as:

f(x) = (x - 4)(x - i)(x + i)

Using the same approach as above, the expression becomes:

f(x) = (x - 4)(x - i)(x + i)(x + (-i))
= (x - 4)(x^2 + xi + 3x - 3i)
= (x - 4)(x^2 + 3x + xi - 3i)

Expanding this equation:

f(x) = (x^3 - 4x^2 + 3xi - 3xi - 12x + 12)(x + i)
= (x^3 - 4x^2 - 12x + 12)(x + i)

To match the given condition f(-3) = 60, we can substitute x = -3 into the expression:

f(-3) = (-3^3 - 4(-3)^2 - 12(-3) + 12)(-3 + i)
= (-27 - 36 + 36 + 12)(-3 + i)
= (-15)(-3 + i)
= 45 - 15i

Hence, your answer of f(x) = 6x^3 + 24x^2 + 6x + 24 is incorrect.

Therefore, both of your answers are incorrect.

To check your answers, we need to substitute the given values into the polynomial functions you provided and see if they satisfy the conditions.

1. For n=3; 3 and i are zeros; f(2)=20
To check your answer, let's substitute the given values into the polynomial f(x) = -4x^3 + 12x^2 - 4x + 12:
f(2) = -4(2)^3 + 12(2)^2 - 4(2) + 12
= -4(8) + 12(4) - 8 + 12
= -32 + 48 - 8 + 12
= 20

Since f(2) is equal to the desired value of 20, your answer is correct.

2. For n=3; 4 and i are zeros; f(-3)=60
To check your answer, let's substitute the given values into the polynomial f(x) = 6x^3 + 24x^2 + 6x + 24:
f(-3) = 6(-3)^3 + 24(-3)^2 + 6(-3) + 24
= 6(-27) + 24(9) - 18 + 24
= -162 + 216 - 18 + 24
= 60

Since f(-3) is equal to the desired value of 60, your answer is correct.

Therefore, both of your answers are correct based on the given conditions.

the first one is correct, the 2nd is not

try f(4) in your equation, you do not get zero.

I had f(x) = a(x^2+1)(x-4)
then for f(-3)=60
60 = a(10)(-7)
a= -6/7

so f(x) = -6/7(x^2+1)(x-4)

expand it for your answer