How do you write a polynomial in standard form given zeros 1,5,3-i?

1 Answer
Apr 14, 2016

Answer:

#x^4-12x^3+51x^2-70x+50=0#.

Explanation:

Complex roots occur in conjugate pairs. So, as #3-i# is a root, #3+i# is also a root.

The roots are 1, 5, #3-i and 3+i#.

The biquadratic having these as roots is in the form
#x^4-S_1 x^3 + S_2 x^2-S_3x+S_4=0#, where
#S_1=sum#(roots) = 12,
#S_2=sum#(products of the roots, taken two at a time) = 51,
#S_3=sum#(product of the roots, taken three at a time) = 70
and #S_4=#product of all the roots =50.

Alternative method:

The factors of the polynomial are #x-1, x - 5, ((x-(3-i)) and (x-(3+i))#.
The biquadratic equation is

#(x-1)(x-3)((x-3)^2+1)=0#.

Expand and write the terms, in the order of descending powers of x.
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