Write a polynomial in standard form , with a zero 3i, an x-intercept 6, a multiplicity 2, and a leading coefficient 1?
Complex roots appear in conjugate form, so we have roots 3i and -3i
giving us two factors of (x-3i)(x+3i) = x^2 + 9
an x-intercept of 6 with multiplicity of 2 yields (x-6)(x-6)
I would state the polynomial as
y = (x^2 + 9)(x-6)^2 , expanding this would be superfluous.
Thanks so much
To write a polynomial with the given specifications, we can start by using the fact that one zero is given as 3i. Since complex zeros occur in conjugate pairs, the other complex zero will be -3i.
We also know that there is an x-intercept at 6, which means that (x - 6) is a factor of the polynomial.
Furthermore, it is given that the multiplicity of the x-intercept 6 is 2. This means that (x - 6)² is a factor of the polynomial.
Finally, it is mentioned that the leading coefficient is 1.
To find the complete polynomial, we can multiply the factors mentioned above:
(x - 3i)(x + 3i)(x - 6)²
Expanding this expression, we get:
(x² - (3i)²)(x - 6)²
Simplifying further:
(x² + 9)(x - 6)²
Multiplying the expression by the leading coefficient of 1:
1(x² + 9)(x - 6)²
Therefore, the polynomial in standard form with the given specifications is:
f(x) = (x² + 9)(x - 6)²