How do you write a polynomial in standard form given the zeros x=7, -13, and 3+2i?

1 Answer
Oct 2, 2016

Answer:

#"The Reqd. Poly."=k(x^4-114x^2+624x-1183), k in CC-{0}.#

Explanation:

Let #p(x)# be the reqd. poly.

The Zeroes of #p(x)" are "7,-13, and, 3+2i#.

We know that the complex zero of a poly. occurs in conjugate pairs.

So, the conjugate of #3+2i, i.e, 3-2i# is also a zero of #p(x).#

In all, thus, we have #4# zeroes of #p(x)#, meaning that #p(x)# must

be of degree #4#, having factors, derived using the zeroes,

#(x-7), (x+13), (x-3-2i), & (x-3+2i).#

We conclude that,

#p(x)=k(x-7)(x+13)(x-3-2i)(x-3+2i), k in CC-{0}#

#=k(x^2+6x-91){(x-3)^2-(2i)^2}#

#=k(x^2+6x-91){x^2-6x+9-4i^2}#

#=k(x^2+6x-91){x^2-6x+9-4(-1)}#

#=k(x^2+6x-91)(x^2-6x+13)#

#=-k{6x+(x^2-91)}{6x-(x^2+13)}#

#=-k{(6x)^2+6x(x^2-91-x^2-13)-(x^2-91)(x^2+13)}#

#=-k{36x^2-624x-(x^4-78x^2-1183)}#

#-k(114x^2-624x-x^4+1183)#

#:. p(x)=k(x^4-114x^2+624x-1183), k in CC-{0}.#