degree 4
zeros i & (1+i)
constant term 12
How do I start this problem.
Thanks
A degree four polynomial will have the form (x^4 +... x + c), where c is a constant. You will need to generate an equation that has the above form, using c=12. Solve for the roots [i & (1+i)].
The equation would be:
x^4 + ax^3 + bx^2 + cx + 12 = 0
To solve for the coefficients a, b, and c, you can use the following equations:
a = -(i + (1+i))
b = i(1+i)
c = -12
Therefore, the polynomial is:
x^4 - (i + (1+i))x^3 + i(1+i)x^2 - 12x + 12 = 0
To start the problem, you can write the equation in the form of a degree four polynomial using the given information.
Let's assume the polynomial is of the form:
f(x) = a(x - r1)(x - r2)(x - r3)(x - r4) + c
Where r1, r2, r3, and r4 are the roots of the polynomial, and c is the constant term.
Based on the given information, we know that the roots of the polynomial are i and (1 + i). This means that we can rewrite the equation as:
f(x) = a(x - i)(x - (1 + i))(x - r3)(x - r4) + 12
Now, we need to find the values of r3 and r4. Since we have a degree four polynomial, there will be two additional roots. We can use the fact that complex roots always come in conjugate pairs to determine the values of r3 and r4.
The complex conjugate of i is -i, and the complex conjugate of (1 + i) is (1 - i). So, we have:
f(x) = a(x - i)(x - (1 + i))(x + i)(x - (1 - i)) + 12
Expanding this equation gives:
f(x) = a(x^2 - x - xi + i)(x^2 - x + xi + i) + 12
Simplifying further gives:
f(x) = a(x^4 - 2x^3 + 2ix^2 + 4x - 3xi + 2i) + 12
Now, equating the coefficient of the x^4 term to zero, we have:
a = 0
So, the equation becomes:
f(x) = 0(x^4 - 2x^3 + 2ix^2 + 4x - 3xi + 2i) + 12
Finally, we can simplify the equation to:
f(x) = 12
Therefore, the polynomial with degree four, the roots i and (1 + i), and the constant term 12 is:
f(x) = 12
To start solving this problem, we can use the fact that complex roots occur in conjugate pairs.
Let's first consider the root i. Complex roots always come in pairs, so the conjugate of i is -i. Therefore, the factors for the roots i and -i are (x-i) and (x-(-i)) which simplify to (x-i) and (x+i), respectively.
Next, let's look at the root 1+i. Once again, we consider the conjugate of 1+i, which is 1-i. The factors for the roots 1+i and 1-i are (x-(1+i)) and (x-(1-i)). Simplifying these factors, we get (x-1-i) and (x-1+i), respectively.
To find the equation with the given roots, we multiply all the factors together:
(x-i)(x+i)(x-1-i)(x-1+i)
By expanding this equation, we can simplify it into a form where we collect like terms and find the expression in the form of (x^4 +... x + c):
(x-i)(x+i)(x-1-i)(x-1+i) = (x^2+ix-ix-i^2)(x^2-x-ix+i)
= (x^2+i^2)(x^2-x)
Now, let's simplify this further:
(x^2+i^2)(x^2-x) = (x^2+(-1))(x^2-x) = (x^2-1)(x^2-x)
= x^4 - x^3 - x^2 + x
The constant term in this equation is 0, not 12 as mentioned in the question. Therefore, it seems there might be a mistake in the given information. Please double-check the question and provide the correct constant term if available.