How do you construct polynomial equations with the given roots?

1. #2#, #4# and #-7#.
2. #5# and #sqrt(3)#.

1 Answer
George C.
Aug 11, 2017

1. #x^3+x^2-34x+56 = 0#

2. #x^3-5x^2-3x+15 = 0#

Explanation:

Note that if a polynomial in #x# has a zero #a# then it has a factor #(x-a)# and vice versa.

#color(white)()#
For question 1 we can construct a polynomial:

#f(x) = (x-2)(x-4)(x+7) = x^3+x^2-34x+56#

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this #f(x)#.

So the polynomial equation:

#x^3+x^2-34x+56 = 0#

has roots #2#, #4# and #-7#.

#color(white)()#
For question 2 we can multiply out #(x-5)(x-sqrt(3))# but this will result in a polynomial with irrational coefficients:

#(x-5)(x-sqrt(3)) = x^2-(5+sqrt(3))x+5sqrt(3)#

If - as is probably the case - we want a polynomial with integer coefficients, then we also need the rational conjugate #-sqrt(3)# to be a zero and #(x+sqrt(3))# a factor.

Then we can define:

#g(x) = (x-5)(x-sqrt(3))(x+sqrt(3)) = (x-5)(x^2-3) = x^3-5x^2-3x+15#

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this #g(x)#.

So the polynomial equation:

#x^3-5x^2-3x+15 = 0#

has roots #5#, #sqrt(3)# and #-sqrt(3)#.

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