Find a degree 3 polynomial with real coefficients having zeros
2 and 4-3i and a lead coefficient of 1.
Why did the polynomial go to the party? Because it wanted to find its real and complex roots!
To find a degree 3 polynomial with real coefficients and zeros 2 and 4-3i, we also know that complex zeros come in conjugate pairs. So, the other complex zero is the conjugate of 4-3i, which is 4+3i.
Therefore, the polynomial is:
(x - 2)(x - (4-3i))(x - (4+3i))
Expanding this, we get:
(x - 2)(x - 4 + 3i)(x - 4 - 3i)
Simplifying further, we have:
(x - 2)(x^2 - 8x + 16 + 9)
(x^3 - 10x^2 + 25x - 2x^2 + 16x - 40 + 9x - 72)
Combining like terms, we finally get:
x^3 - 12x^2 + 50x - 103
So, the degree 3 polynomial with real coefficients having zeros 2 and 4-3i is: x^3 - 12x^2 + 50x - 103.
To find the degree 3 polynomial with real coefficients having zeros 2 and 4-3i, we can use the fact that complex conjugates always come in pairs.
Since 4-3i is a zero, its conjugate, 4+3i, must also be a zero.
The polynomial can be written as:
(x - 2)(x - (4-3i))(x - (4+3i))
Now let's simplify this polynomial:
(x - 2)(x - 4 + 3i)(x - 4 - 3i)
Expanding this expression gives us:
(x - 2)(x^2 - 4x + 3ix - 4x + 16 - 12i - 3ix + 12i - 9i^2)
Simplifying further:
(x - 2)(x^2 - 8x + 16 + 9)
(x^3 - 8x^2 + 16x + 9x - 18)
Finally, we can write the polynomial in standard form:
x^3 - 8x^2 + 25x - 18
Therefore, the degree 3 polynomial with real coefficients having zeros 2 and 4-3i and a lead coefficient of 1 is:
f(x) = x^3 - 8x^2 + 25x - 18
To find a degree 3 polynomial with real coefficients having the given zeros, we need to consider the complex conjugate of the given complex zero. The complex conjugate of 4-3i is 4+3i.
A polynomial with the given zeros can be written in factored form as:
(x - 2)(x - (4-3i))(x - (4+3i))
Simplifying this gives:
(x - 2)(x - 4 + 3i)(x - 4 - 3i)
= (x - 2)((x - 4) + 3i)((x - 4) - 3i)
= (x - 2)((x - 4)^2 - (3i)^2)
= (x - 2)((x - 4)^2 + 9)
= (x - 2)(x^2 - 8x + 16 + 9)
= (x - 2)(x^2 - 8x + 25)
Finally, since the leading coefficient is 1, the desired polynomial is:
f(x) = (x - 2)(x^2 - 8x + 25)
if the coefficients are real, then the complex roots occur in pairs. So, our polynomial is
(x-2)(x-(4-3i))(x-(4+3i))
(x-2)((x-4)+3i)((x-4)-3i)
(x-2)((x-4)^2+3^2)
...