Find the polynomial of degree #4# having zeros: #8#(multiplicity 2), #sqrt11# and #-sqrt11#?

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Answer 1 · 2017-11-15T07:29:13

Polynomial is #x^4-16x^2+53x+704#

A polynomial of degree #4# having zeros #a,b,c# and #d# is

#(x-a)(x-b)(x-c)(x-d)#, which can be expanded as

#x^4-(a+b+c+d)x^2+(ab+bc+cd+ac+bd)x-abcd#

If we have #a# (multiplicity #2# and #b# and #c#, then it would be

#(x-a)^2(x-b)(x-c)#

Hence in a polynomial of degree #4# having zeros: #8#(multiplicity 2), #sqrt11,-sqrt11 #, the polynomial would be

#(x-8)^2(x-sqrt11)(x+sqrt11)#

= #x^4-16x^2+(64+8sqrt11-8sqrt11+8sqrt11-8sqrt11-11)x-8xx8xx(-11)#

= #x^4-16x^2+53x+704#