Question #528b9

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Answer 1 · 2018-01-21T18:58:33

#f(x)=x^4-14x^3+87x^2-200x+208#

we are given

roots

#4#(multiplicity 2)

#3+2i#

now since the polynomial has real coefficients complex roots appear in conjugate pairs

so another root is #3-2i#

we can write the polynomial as:

#f(x)=(x-(3+2i))(x-(3-2i))(x-4)^2#

we now multiply and simplify

#f(x)=(x-3-2i)(x-3+2i)(x^2-8x+16)#

#f(x)=((x-3)^2-(2i)^2)(x^2-8x+16)#

#f(x)=(x^2-6x+9+4)(x^2-8x+16)#

#f(x)=(x^2-6x+13)(x^2-8x+16)#

#=x^4-8x^3+16x^2#

#color(white)(aaaaa)-6x^3+48x^2-96x#

#color(white)(aaaaaaaaaa)+13x^2-104x+208#

#f(x)=x^4-14x^3+87x^2-200x+208#