How many times does the graph of f(x)=x^3+3x^2+3x+1 touch the x axis?

Question

Please explain

Answer 1

This is another way of asking you to find the "zeros" or where y = 0. so replace f(x) with zero and solve the polynomial.

Once factored...
(x+1)(x+1)(x+1) = 0 if any of the parenthesis is zero on the inside, then the whole thing will be zero so x +1 = 0

y= 0 at x = -1. It only crosses once since they are all the same.

Answer 2

(x+1)(x+1)(x+1)=f(x) so there is only 1 place where this function touches the x-axis.

Answer 3

f(x)=x^3+3x^2+3x+1
=x^3+3x^2+3x+1
=x^3+1+3x^2+3x
=(x+1)(x^2-x+1)+3x(x+1)
=(x+1)(x^2-x+1+3x)
=(x+1)(x^2+2x+1)
=(x+1)^3
f(x)=0
x=-1
one time touch the x axis

Answer 4

factor it. and find the real roots. the number of real roots = number of times it crosses the axis. also, you may graph it on the calculator, in which case you find out that it crosses once.