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
[tex]p(x)=(x+2)(x+4)(x+3-4i)(x+3+4i)[/tex]
[tex]p(x)=(x^2+4x+2x+8)((x+3)^2-(4i)^2)[/tex] [tex][\because a^2-b^2=(a-b)(a+b)][/tex]
[tex]p(x)=(x^2+6x+8)(x^2+6x+9-16i^2)[/tex] [tex][\because (a+b)^2=a^2+2ab+b^2][/tex]
[tex]p(x)=(x^2+6x+8)(x^2+6x+9-16(-1))[/tex] [tex][\because i^2=-1][/tex]
[tex]p(x)=(x^2+6x+8)(x^2+6x+9+16)[/tex]
[tex]p(x)=(x^2+6x+8)(x^2+6x+25)[/tex]
[tex]p(x)=x^2(x^2+6x+25)+6x(x^2+6x+25)+8(x^2+6x+25)[/tex]
[tex]p(x)=x^4+6x^3+25x^2+6x^3+36x^2+150x+8x^2+48x+200[/tex]
Combining like terms, we get
[tex]p(x)=x^4+(6x^3+6x^3)+(25x^2+36x^2+8x^2)+(150x+48x)+200[/tex]
[tex]p(x)=x^4+12x^3+69x^2+198x+200[/tex]
Therefore, the required polynomial is [tex]p(x)=x^4+12x^3+69x^2+198x+200[/tex].
