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olganol [36]
2 years ago
7

Degree 3 polynomial with integer coefficients with zeros -2i and 7/5

Mathematics
1 answer:
pav-90 [236]2 years ago
8 0

Answer:

P(x)=5x^3-7x^2+20x-28

Step-by-step explanation:

<u>Polynomial Construction</u>

A polynomial can be constructed by knowing its roots or zeros and the leading coefficient.

In our example, we have only two roots and no leading coefficient, thus the solution provided will be only one of the infinite possible solutions.

There is one imaginary solution x=-2i, we know it the polynomial has integer coefficients, then the imaginary roots come in conjugated pairs, meaning the other root must be x=+2i. Having completed all the roots, our polynomial is

P(x)=(x-2i)(x+2i)(x-7/5)

Operating

P(x)=(x^2+4)(5x-7)/5

We can discard the denominator since the roots of the polynomial will be the same with or without it, and we need the coefficients to be integer.

Operating

P(x)=5x^3+20x-7x^2-28

Rearranging

\boxed{P(x)=5x^3-7x^2+20x-28}

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