Find the polynomial function f with Real coefficients that has the given degree 3 zeros -2, 1-sqrt 2i, and solution point -42
since complex roots come in conjugate pairs,
f(x) = a(x+2)(x-(1-i√2))(x-(1+i√2))
= a(x+2)((x-1)^2+2)
Not sure what a "solution point" is, since you have already given the roots.
But if f(0) = 42, then
a(2)(1+2) = 42
a = 7
and f(x) = 7(x+2)(x^2-2x+3)
To find the polynomial function, we can use the fact that the zeros of a polynomial are the values that make the polynomial equal to zero.
Given that the polynomial has degree 3 and the zeros are -2, 1 - sqrt(2)i, and a solution point is (-42, 0), we can set up the following equation:
f(x) = a(x - (-2))(x - (1 - sqrt(2)i))(x - c)
Where 'a' is a constant that scales the polynomial and 'c' is the remaining zero.
To find 'c', we can use the solution point (-42, 0):
f(-42) = a((-42) - (-2))((-42) - (1 - sqrt(2)i))((-42) - c) = 0
Simplifying this equation, we get:
(a)(40)(42 + 1 - sqrt(2)i)(42 + 2) = 0
Expanding and multiplying, we get:
(a)(40)(44 + 42 - sqrt(2)i - sqrt(2)i) = 0
(a)(176)(86 - 2sqrt(2)i) = 0
This equation implies that either 'a' = 0 or (86 - 2sqrt(2)i) = 0.
If 'a' = 0, then the polynomial function would be f(x) = 0, which is not the desired polynomial.
Therefore, we need to solve for 'c' in the equation (86 - 2sqrt(2)i) = 0:
86 - 2sqrt(2)i = 0
2sqrt(2)i = 86
sqrt(2)i = 43
Thus, c = 43.
Now, we have the zeros: -2, 1 - sqrt(2)i, and 43.
Substituting these values back into the equation, we have:
f(x) = a(x + 2)(x - 1 + sqrt(2)i)(x - 43)
Now we can use the solution point (-42, 0) to find the value of 'a':
f(-42) = a((-42) + 2)((-42) - 1 + sqrt(2)i)((-42) - 43) = 0
(a)(-40)(-43 - 1 + sqrt(2)i)(-85) = 0
Expanding and multiplying, we get:
(a)(-40)(-44 + sqrt(2)i)(-85) = 0
(a)(-40)(3760 - 85sqrt(2)i) = 0
This equation implies that either 'a' = 0 or (3760 - 85sqrt(2)i) = 0.
If 'a' = 0, then the polynomial function would be f(x) = 0, which is not the desired polynomial.
Therefore, we need to solve for 'a' in the equation (3760 - 85sqrt(2)i) = 0:
3760 - 85sqrt(2)i = 0
85sqrt(2)i = 3760
sqrt(2)i = 44
Thus, a = -40.
Finally, substituting 'a' and 'c' back into the polynomial equation, we have:
f(x) = -40(x + 2)(x - 1 + sqrt(2)i)(x - 43)
Therefore, the polynomial function f with real coefficients that has degree 3, zeros -2, 1 - sqrt(2)i, and a solution point (-42, 0) is:
f(x) = -40(x + 2)(x - 1 + sqrt(2)i)(x - 43)
To find the polynomial function f, we will use the fact that a polynomial of degree n with real coefficients can be written as:
f(x) = a_n * (x - r_1) * (x - r_2) * ... * (x - r_n)
where a_n is the leading coefficient, r_1, r_2, ..., r_n are the roots of the polynomial.
In this case, we are given the degree 3 zeros -2, 1 - sqrt(2)i, and the solution point (-42, 0). Let's break down the steps to find the polynomial function f:
1. Degree 3 zeros:
We are given two zeros, -2 and 1 - sqrt(2)i. Since the coefficient of a polynomial with real coefficients must be real, the complex conjugate of 1 - sqrt(2)i, which is 1 + sqrt(2)i, must also be a zero. So, the three zeros are -2, 1 - sqrt(2)i, and 1 + sqrt(2)i.
2. Solution point:
We are given the solution point (-42, 0). The x-coordinate of the solution point is -42, which means that f(-42) = 0.
3. Constructing the polynomial:
Using the zeros and the solution point, we can construct the polynomial function as follows:
f(x) = a_n * (x - r_1) * (x - r_2) * (x - r_3)
where r_1 = -2, r_2 = 1 - sqrt(2)i, and r_3 = 1 + sqrt(2)i.
Since f(-42) = 0, we can substitute x = -42 into the equation:
0 = a_n * (-42 - r_1) * (-42 - r_2) * (-42 - r_3)
Now, let's plug in the values of the zeros:
0 = a_n * (-42 - (-2)) * (-42 - (1 - sqrt(2)i)) * (-42 - (1 + sqrt(2)i))
Simplifying the expression:
0 = a_n * (-42 + 2) * (-42 - 1 + sqrt(2)i) * (-42 - 1 - sqrt(2)i)
0 = a_n * (-40) * (-84 + sqrt(2)i) * (-84 - sqrt(2)i)
To find the leading coefficient, a_n, we can divide both sides by the product of the terms in brackets:
0 = a_n * (-40) * (-84 + sqrt(2)i) * (-84 - sqrt(2)i) / (-40) * (-84 + sqrt(2)i) * (-84 - sqrt(2)i)
Cancelling out the common factors:
0 = a_n * (-84 + sqrt(2)i) * (-84 - sqrt(2)i)
Since the expression on the right-hand side is a product equal to zero, at least one of its factors must be zero. Hence, we have:
a_n = 0
or
(-84 + √2i) = 0
or
(-84 - √2i) = 0
Since the leading coefficient a_n cannot be zero for a non-zero polynomial, we have:
(-84 + √2i) = 0
or
(-84 - √2i) = 0
Solving these equations, we find:
√2i = 84
or
√2i = -84
Squaring both sides to eliminate the square root:
2i = 84^2
or
2i = (-84)^2
Simplifying both equations:
2i = 7056
or
2i = 7056
Since the imaginary number i is defined as the square root of -1, the equations have no real solutions.
Therefore, there is no polynomial function f with real coefficients that has the given degree 3 zeros -2, 1 - sqrt(2)i, and solution point -42.