2,3,i√13
find the polynomial equation
assuming those are the roots, and remembering that the complex roots come in conjugate pairs,
f(x) = (x-2)(x-3)(x-i√13)(x+i√13)
= (x-2)(x-3)(x^2+13)
with the real coefficients that has given roots
To find the polynomial equation with roots 2, 3, and i√13, we can use the fact that for a polynomial equation, the roots are the values of x that make the equation equal to zero.
Since 2 and 3 are real numbers, their conjugates are also roots of the polynomial equation. Therefore, the conjugates of 2 and 3 are -2 and -3, respectively.
To find the equation, we start by considering the factors for the roots. The factor for x = 2 is (x - 2), and the factor for x = 3 is (x - 3). Similarly, the factor for x = -2 is (x + 2), and the factor for x = -3 is (x + 3).
Since i√13 is a complex number, its conjugate is also a root. The conjugate of i√13 is -i√13. Therefore, the factor for x = i√13 is (x - i√13) and the factor for x = -i√13 is (x + i√13).
Now, we can write the polynomial equation as:
(x - 2)(x - 3)(x + 2)(x + 3)(x - i√13)(x + i√13) = 0
Multiplying these factors together, we get:
(x² - 4)(x² - 9)(x² + 13) = 0
Expanding this equation, we get:
(x⁶ - 4x⁴ - 9x⁴ + 36x² + 13x⁴ - 52x² - 117x² + 468) = 0
Combining like terms, we have:
x⁶ - 13x⁴ + 25x² + 468= 0
Therefore, the polynomial equation is:
x⁶ - 13x⁴ + 25x² + 468 = 0