How do you find the zeros with multiplicity for the function #p(x)=(x^3-8)(x^5-4x^3)#?

1 Answer
Nov 15, 2016

Answer:

#p(x)# has zeros:

#0# with multiplicity #3#

#2# with multiplicity #2#

#-2# with multiplicity #1#

#-1+sqrt(3)i# with multiplicity #1#

#-1-sqrt(3)i# with multiplicity #1#

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

We find:

#p(x) = (x^3-8)(x^5-4x^3)#

#color(white)(p(x)) = (x^3-2^3)x^3(x^2-2^2)#

#color(white)(p(x)) = (x-2)(x^2+2x+4)x^3(x-2)(x+2)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)(x^2+2x+4)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)(x^2+2x+1+3)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)((x+1)^2-(sqrt(3)i)^2)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)#

#color(white)(p(x)) = x^3(x-2)^2(x+2)(x+1-sqrt(3)i)(x+1+sqrt(3)i)#

Hence zeros:

#0# with multiplicity #3#

#2# with multiplicity #2#

#-2# with multiplicity #1#

#-1+sqrt(3)i# with multiplicity #1#

#-1-sqrt(3)i# with multiplicity #1#