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OverLord2011 [107]
4 years ago
7

Help needed quick and fast

Mathematics
1 answer:
Maksim231197 [3]4 years ago
6 0

Answer:

rational : square root of 81, 7/16, square root of 36

irrational : pi, square root of 48, square root of 120

Step-by-step explanation:

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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 an
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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.

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3 years ago
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