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Construet a polynomial with integer coefficients for which 2+i is a zero.

User Gamlor
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We know that if a polynomial with real coefficients has a complex root, then its conjugate must also be a root. This stems from the Fundamental Theorem of Algebra, which implies that non-real roots of real-coefficient polynomials must occur in complex conjugate pairs.

Given that 2+i is a root of this polynomial, the conjugate, which would be 2-i, must also be a root.

The fundamental theory of algebra also tells us how to construct a polynomial when we know the roots. If r1, r2, ..., rn are the roots of a polynomial, then the polynomial can be written in the form of:

P(x) = a(x - r1)(x - r2)...(x - rn)

where 'a' is a non-zero constant.

Using what we know, our polynomial can thus be written as:

P(x) = a(x - (2 + i))(x - (2 - i))

where 'a' is a real number we can choose for simplicity. Let's choose 'a' to be 1.

Next, we expand the polynomial:

P(x) = (x - 2 - i)(x - 2 + i)

This simplification uses the property of complex numbers which is (a + bi)(a - bi) = a² - (bi)². In simpler terms, this is equal to the difference of two squares, which are (x - 2)² - (i)².

This simplifies to:

P(x) = x² - 4x + 4 - (-1)

Solving this, we have:

P(x) = x² - 4x + 5

So, our resultant polynomial is x² - 4x + 5. This polynomial satisfies the conditions that it has integer coefficients and that one of its zeros is 2 + i.

The Polynomial constructed using the integer coefficients for which 2+i is a zero is x² - 4x + 5

User Youjun Hu
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