Question

Find the polynomial of minimum degree, with real coefficients, zeros at x=4±2⋅i and x=−8, and y-intercept at −480. Write your answer in standard form.

Answer 1

The polynomial is found by creating factors from the given zeros (both the real zero and the conjugate pair of complex zeros), multiplying them, and scaling the result to achieve the specified y-intercept.

Step-by-step explanation:

The question is asking us to find the polynomial of minimum degree with real coefficients that has specific zeros and a y-intercept. Since complex zeros in polynomials with real coefficients come in conjugate pairs, the zeros at x=4+2i and x=4-2i must be included. Additionally, we have the zero at x=-8. To find the polynomial, we multiply the factors associated with these zeros: (x-(4+2i))(x-(4-2i))(x-(-8)). Expanding this and simplifying will give us a cubic polynomial.

To ensure the y-intercept is at -480, we need to multiply our polynomial by a constant that will scale the y-intercept accordingly. If we set x=0 in the polynomial, the value we obtain is the y-intercept. To get -480, we will have to adjust the leading coefficient of the polynomial so that when x=0, the polynomial evaluates to -480.

Answer 2

The polynomial of minimum degree is
`p(x) = x^(4)-3\cdot x^(3)-44\cdot x^(2)+292\cdot x-480`.

Step-by-step explanation:

According to the statement, we appreciate that polynomial pass through the following points:

(i)
`(4 + 2\,i,0)`, (ii)
`(4 - 2\,i, 0)`, (iii)
`(-8, 0)`, (iv)
`(0, -480)`

From Algebra we know that any n-th grade polynomial can be constructed by knowing n+1 different points. Hence, the polynomial of minimum degree is a quartic function. The polynomial has the following form:

`p(x) = (x-4-2\,i)\cdot (x-4+2\,i)\cdot (x+8)\cdot (x-r_(1))`

We proceed to expand the expression until standard form is obtained:

`p(x) = (x^(2)-4\cdot x-2\,i\cdot x-4\cdot x +16+8\,i+2\,i\cdot x-8\,i+4)\cdot (x+8)\cdot (x-r_(1))`

`p(x) = (x^(2)-8\cdot x+20)\cdot (x+8)\cdot (x-r_(1))`

`p(x) = (x^(3)-8\cdot x^(2)+20\cdot x +8\cdot x^(2)-64\cdot x+160)\cdot (x-r_(1))`

`p(x) = (x^(3)-44\cdot x +160)\cdot (x-r_(1))`

If we know that
`p (0) = -480`, then:

`-480=160\cdot (-r_(1))`

`-160\cdot r_(1) = -480`

`r_(1) = 3`

Then, the polynomial is:

`p(x) = (x^(3)-44\cdot x +160)\cdot (x-3)`

`p(x) = x^(4)-44\cdot x^(2)+160\cdot x-3\cdot x^(3)+132\cdot x -480`

`p(x) = x^(4)-3\cdot x^(3)-44\cdot x^(2)+292\cdot x-480`

The polynomial of minimum degree is
`p(x) = x^(4)-3\cdot x^(3)-44\cdot x^(2)+292\cdot x-480`.