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

2 Answers
May 4, 2016

Answer:

#x=-2, 1, -1#

Explanation:

So, I like to factor this sort of problm using synthetic division. First thing's first, let's set up this problem.

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#
#color(white)(-2)|#
#-2|#__________

We bring down the #1#, which gives us this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#
#color(white)(-2)|#
#-2|#__________
#color(white)(0)color(white)(000)1#

We then multiply the #-2# by the #1# that we brought down to give us #-2#. That value is brought up to the next row, which brings us to this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#
#color(white)(-2)|##color(white)(00)-2#
#-2|#__________
#color(white)(0)color(white)(000)1#

From here, we just add the #2# to the #-2# and end up at this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#
#color(white)(-2)|##color(white)(00)-2#
#-2|#__________
#color(white)(0)color(white)(000)1color(white)(000)0#

If we continue this system, we end up with this:

#color(white)(-2)|##1color(white)(000)2color(white)(000)-1color(white)(000)-2#
#color(white)(-2)|##color(white)(00)-2color(white)(0000)0color(white)(000000)2#
#-2|#________
#color(white)(0)color(white)(000)1color(white)(000)0color(white)(00)-1color(white)(000000)0#

So now we have #x=-2# or #x+2=0#, or #(x+2)# and #(x^2-1)#. This can be simplified to just #(x+1)(x-1)#.

We now have the factors of #x^3+2x^2-x-2#, which are #(x+2)(x+1)(x-1)#, which can be rewritten as: #x=-2,-1,1#.

So let's just double check that we found all the zeros of #x^3+2x^2-x-2# be graphing it

graph{x^3+2x^2-x-2}

And look, the three #x#-intercepts are what we said, #x=1,-1,-2#.

Jun 14, 2016

Answer:

The zeros are #-2, -1, 1.#

Explanation:

Method 1

#f(x)=x^3+2x^2-x-2=x^2(x+2)-1(x+2)=(x+2)(x^2-1)=(x+2)(x+1)(x-1).#

Hence, the zeros are #-2, -1, 1.#

Method 2

Observe that the sum of the co-effs. #=1+2-1-2=0#

So, #(x-1)# is a factor of poly. #f(x)#. Now, we will rearrange the terms of the poly. in such a way that #(x-1)# can be taken out as a common factor. See the following computation :-

#f(x)=x^3+color(red)2x^2-color(blue)1x-2=x^3-color(red)1x^2+color(red)3x^2-color(blue)3x+color(blue)2x-2=x^2(x-1)+3x(x-1)+2(x-1)=(x-1)(x^2+3x+2)=(x-1){x^2+(2+1)x+(2xx1)}=(x-1){x^2+2x+x+2}=(x-1){x(x+2)+1(x+2)}=(x-1)(x+2)(x+1).#

Hence, the zeros are #1, -2, -1# as before!