Question #802a8

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Answer 1 · 2017-05-30T15:06:11

The polynomial is #f(x)=x^3+6x+20#

We need

#(a+b)(a-b)=a^2-b^2#

If one root is #=1-3i# then the conjugate is also a root #=1+3i#

#(1+3i)(1-3i)=1-(9i^2)=10#

as

#i^2=-1#

So,

#f(x)=(x-(1-3i))(x-(1+3i))(x+2)#

#f(x)=(x^2-x(1+3i)-x(1-3i)+(1+3i)(1-3i))(x+2)#

#=(x^2-x-3ix-x+3ix+10)(x+2)#

#=(x^2-2x+10)(x+2)#

#=x^3+2x^2-2x^2-4x+10x+20#

#=x^3+6x+20#