What are the zeros of the function? What are their multiplicities? f(x)=x^4-4x+3x^2

1 Answer
Nov 25, 2017

Answer:

#x=0,x=1" and "x=-1/2+-sqrt15/2i#

Explanation:

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

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

#"the coefficients of "x^3+3x-4#

#1+3-4=0rArrx=1" is a root and "(x-1)" is a factor"#

#color(red)(x^2)(x-1)color(magenta)(+x^2)+3x-4#

#=color(red)(x^2)(x-1)color(red)(+x)(x-1)color(magenta)(+x)+3x-4#

#=color(red)(x^2)(x-1)color(red)(+x)(x-1)color(red)(+4)(x-1)cancel(color(magenta)(+4))cancel(-4)#

#=color(red)(x^2)(x-1)color(red)(+x)(x-1)color(red)(+4)(x-1)+0#

#rArrx(x-1)(x^2+x+4)=0#

#rArrx=0,x=1to("with multiplicity of 1")#

#"check the "color(blue)"discriminant "" of "x^2+x+4#

#"with "a=1,b=1,c=4#

#Delta=b^2-4ac=1-16=-15#

#"indicating there are 2 complex roots"#

#"using the "color(blue)"quadratic formula"#

#x=(-1+-sqrt(-15))/2#

#rArrx=-1/2+-sqrt15/2i#

#"there are 2 real zeros and 2 complex zeros"#