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

Answer:

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 $p(x) = x^{4}-3\cdot x^{3}-44\cdot x^{2}+292\cdot x-480$.