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Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. –2, –4, –3 + 4i

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Given:

Zeroes of a polynomial are –2, –4, –3 + 4i.

To find:

The polynomial function of least degree with real coefficients in standard form.

Solution:

According to the complex conjugate root theorem, if a complex number a+ib is a zero of a polynomial, then its conjugate a-ib is also a zero of than polynomial.

–3 + 4i is zero of the polynomial. So, by complex conjugate root theorem -3-4i is also a zero of required polynomial.

If c is a zero of p(x), then (x-c) is a factor of p(x).

–2, –4, –3 + 4i, -3-4i are zeroes of the polynomials. So, (x+2), (x+4), (x+3-4i), (x+3+4i) are the factors of the required polynomial.

Let the required polynomial be p(x), so


p(x)=(x+2)(x+4)(x+3-4i)(x+3+4i)


p(x)=(x^2+4x+2x+8)((x+3)^2-(4i)^2)
[\because a^2-b^2=(a-b)(a+b)]


p(x)=(x^2+6x+8)(x^2+6x+9-16i^2)
[\because (a+b)^2=a^2+2ab+b^2]


p(x)=(x^2+6x+8)(x^2+6x+9-16(-1))
[\because i^2=-1]


p(x)=(x^2+6x+8)(x^2+6x+9+16)


p(x)=(x^2+6x+8)(x^2+6x+25)


p(x)=x^2(x^2+6x+25)+6x(x^2+6x+25)+8(x^2+6x+25)


p(x)=x^4+6x^3+25x^2+6x^3+36x^2+150x+8x^2+48x+200

Combining like terms, we get


p(x)=x^4+(6x^3+6x^3)+(25x^2+36x^2+8x^2)+(150x+48x)+200


p(x)=x^4+12x^3+69x^2+198x+200

Therefore, the required polynomial is
p(x)=x^4+12x^3+69x^2+198x+200.

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