Question 118563
You need to use long division or synthetic division to verify that the given factors are indeed factors of the function.  I'll try to walk you through the steps for one of them:


{{{(2x^3+x^2-5x+2)/(x+2)}}}


The first term of the denominator, x, goes into the first term of the numerator, {{{2x^3}}}, {{{2x^2}}} times, so that becomes the first term of your quotient.  {{{2x^2*(x+2)=2x^3+4x^2}}} so subtract these two terms from the first two terms of the numerator expression yielding {{{-3x^2}}}.  Bring down the next numerator term and divide the resulting binomial, {{{-3x^2-5x}}}, by {{{x+2}}}.  The first term of the divisor, x, goes into the first term {{{-3x}}} times, so that becomes the second term of your quotient.  {{{-3x*(x+2)=3x^2-6x}}}.  Subtract these two terms from the ({{{-3x^2-5x}}} dividend part resulting in {{{x}}}.  Bring down the {{{+2}}} resulting in {{{x+2}}}.  Divide the {{{x+2}}} by {{{x+2}}} resulting in 1, which becomes the third term of your quotient.  The quotient is now {{{2x^2-3x+1}}} and there is no remainder.  Since there is no remainder, you have verified that {{{x+2}}} is a factor of {{{(2x^3+x^2-5x+2)}}}.


Repeat the process using the quotient you just derived, {{{2x^2-3x+1}}}, and dividing that by the other given factor, {{{x-1}}}.  Since the process is the same as described above, I'll leave the details to you, but you should get a quotient of {{{2x-1}}} with no remainder.  That tells you that {{{x-1}}} is a factor and that the third factor is {{{2x-1}}}.


Now you can say that {{{f(x)=2x^3+x^2-5x+2=(x+2)(x-1)(2x-1)}}}.  The zeros of f are those values of x for which {{{f(x) = 0}}}.  


{{{f(x) = 0}}} if and only if {{{x+2=0}}} or {{{x-1=0}}} or {{{2x-1=0}}}.


Solving each of these equations yields:


{{{x[1]=-2}}}
{{{x[2]=1}}}
{{{x[3]=1/2}}}


{{{graph(600,600,-8,8,-8,8,2x^3+x^2-5x+2)}}}


Notice that the graph crosses the x-axis at the three zeros of f.


Notice that this graph has a peak and a valley.  It's beyond the scope of this discussion why, but the values of x at the top of the peak and the bottom of the valley are given by solving {{{6x^2+2x-5=0}}}, yielding {{{(-1+-sqrt(31))/6}}}, very roughly speaking -1 and 3/4, as you can see on the graph.  The y values for these points would obviously be the original function evaluated at these two x values.


Hope that helps.
John