How do you list all possible roots and find all factors and zeroes of #x^3+4x^2+5x+2#?

1 Answer
Sep 1, 2016

Answer:

# p(x)=x^3+4x^2+5x+2=(x+1)^2(x+2)#.

The roots of #"p(x)=0, &, the zeroes of p(x) are, "-1,-1,-2#.

Explanation:

Name the given poly. #p(x)=x^3+4x^2+5x+2#.

We notice that,

#"The sum of the co-effs. of odd-powered terms"=1+5=6, &,#

#"The sum of the co-effs. of even-powered terms"=4+2=6.#

This means that #x+1# is a factor of #p(x)#.

# "Now p(x)"=x^3+4x^2+5x+2#,

#=ul(x^3+x^2)+ul(3x^2+3x)+ul(2x+2)#,

#=x^2(x+1)+3x(x+1)+2(x+1)#,

#=(x+1)(x^2+3x+2)#,

#=(x+1){ul(x^2+2x)+ul(x+1)}#,

#=(x+1){x(x+1)+1(x+1)}#,

#=(x+1){(x+1)(x+2)}#,

#=(x+1)^2(x+2)#.

Clearly, the roots of #"p(x)=0, &, the zeroes of p(x) are, "-1,-1,-2#.