F(x)= a( )
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Complex zeroes always come in complementary pairs (conjugate pairs). For degree 5, we would have two pairs of complex zeroes. We have been given one of each pair, so we can easily find the other, which reverses the sign of the imaginary part. So the complement of i is -i, and of 3+i it’s 3-i.

If x is the variable, we have complex factors x-i and x+i, and when we multiply them:


And (x-3-i)(x-3+i)=(x-3)²+1=x²-6x+10.

The real zero gives x+8 as the factor and a polynomial would be (x+8)(x²+1)(x²-6x+10)=

x⁵+2x⁴-37x³+82x²-38x+80. Other suitable polynomials would be multiples of this one:


by Top Rated User (782k points)

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