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