How do you find the polynomial function with roots 1, 7, and -3 of multiplicity 2?

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Answer 1 · 2015-08-10T20:25:21

#f(x)=2(x-1)(x-7)(x+3)=2x^3-5x^2-17x+21#

If the roots are 1,7,-3 then in factored form the polynomial function will be:

#f(x)=A(x-1)(x-7)(x+3)#

Repeat the roots to get the required multiplicity:

#f(x)=(x-1)(x-7)(x+3)(x-1)(x-7)(x+3)#

Answer 2 · 2015-08-11T21:15:39

The simplest polynomial with roots #1#, #7# and #-3#, each with multiplicity #2# is:

#f(x) = (x-1)^2(x-7)^2(x+3)^2#

#=x^6-10x^5-9x^4+212x^3+79x^2-714x+441#

Any polynomial with these roots with at least these multiplicities will be a multiple of #f(x)#, where...

#f(x) = (x-1)^2(x-7)^2(x+3)^2#

#=(x^3-5x^2-17x+21)^2#

#=x^6-10x^5-9x^4+212x^3+79x^2-714x+441#

...at least I think I've multiplied this correctly.

Let's check #f(2)# :

#2^6-10*2^5-9*2^4+212*2^3+79*2^2-714*2+441#

#=64-320-144+1696+316-1428+441=625#

#((2-1)(2-7)(2+3))^2 = (1*-5*5)^2 = (-25)^2 = 625#