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

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.