How do you find all the zeros of #f(x)=2x^3+5x^2+4x+1#?

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Answer 1 · 2016-08-03T22:06:39

#f(x)# has zeros: #-1, -1, -1/2#

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

Note that:

#f(-1) = -2+5-4+1 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#2x^3+5x^2+4x+1#

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

Substituting #x=-1# in the remaining quadratic, we find:

#2x^2+3x+1 = 2-3+1 = 0#

So #x=-1# is a zero again and #(x+1)# a factor:

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

The final zero is #x = -1/2#