Find a polynomial function P(x) having leading coefficient 1, least possible degree, real coeffi

Find a polynomial function P(x) having leading coefficient...

Key Concepts

Complex Conjugate Root Theorem

This theorem states that if a polynomial has real coefficients and a complex number appears as a zero, then its complex conjugate is also a zero. It is therefore essential in determining the complete set of zeros for polynomials with real coefficients when one non-real zero is known.

Multiplicity of Zeros

Multiplicity refers to the number of times a particular zero appears in the factorization of the polynomial. When a zero has a multiplicity greater than one, it means that the corresponding factor appears raised to that power, which affects the shape of the polynomial’s graph at the intercept.

Leading Coefficient and Monic Polynomials

In a monic polynomial, the leading coefficient is 1. This specification is used to uniquely identify the polynomial by ensuring that the highest degree term of the polynomial is unscaled, which simplifies comparisons and constructions of the polynomial.

Polynomial Factorization

Factoring a polynomial involves expressing it as a product of its linear factors corresponding to its zeros. This method is used to construct a polynomial from its zeros, especially when taking into account multiplicities and the need to satisfy real coefficients, thereby ensuring all relevant zeros are incorporated appropriately.

Transcript

00:01 In this problem it is given that the zeros of the required polynomial are 1 plus 2 iota and 2 with a multiplicity of now from conjugate root's theorem we know that if 1 plus 2 iota is a root then its conjugate 1 minus 2 aota is also a root therefore if we list all the roots or zeros we get 1 plus 2 iota 1 minus 2 iota and then 2 comes twice because it has a multiplicity of 2 so now if we write all the factors in this polynomial it is x minus 2 the whole square because 2 has a multiplicity of 2 and then it is x minus 1 plus 2 a iota and x minus 1 minus 2 aota so now if we simply multiply all these 3 factors we will get the required polynomial so let us multiply x minus 2 whole square can be written as x square minus 4x plus 4 and if we multiply these two terms we will get first we'll multiply each term by the corresponding term in the other part so we have x square minus x minus two iota x then multiplying the next term which is minus one to the entire second term we get minus x plus one plus two iota and then multiplying this two iota to the entire second term we get plus two iota x minus two iota minus two iota square oh i'm sorry this is minus four iota square so let us fix that we have minus four iota square now if we see in this entire expression we can cancel out a few terms as they subtract each other off we have minus 2 iota x and plus 2 iota x also we have minus 2 iota and plus 2 iota so finally we get x square minus 4x plus 4 times here we have x square minus 2x now here we have plus 1 and here we have minus 4 iota square we know iota square is minus 1 so we can take this entire term to be plus 4 so here we have plus 5 now if we multiply these two terms again we'll multiply each term of this first part to each corresponding term of the second part so we have x square times x square is x to the power 4 next we have minus 2 x cube and then we have plus 5 x square now we multiply minus 4x to this second term we get minus 4 x cube plus plus 8 x square minus 20 x and then multiplying 4 to this second part we get plus 4 x squared minus 8 x plus 20 so now we can add all the terms with same powers x to the power 4 now we have two terms with x cube so adding them we get minus 6 x squared...

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