Question 106019
{{{P(X)= x3-4x2+x}}}

{{{P(x)}}} = {{{x}}}*({{{x^2}}} – {{{4x}}} + {{{1}}}) this product will be equal to {{{0}}} 
if {{{x=0}}} or  ({{{x^2}}} – {{{4x}}} + {{{1}}}) is equal to {{{0}}}

So,
 One zero will be for {{{x=0}}}

To find other two zeros, we need to solve 
{{{ x^2}}} – {{{4x}}} + {{{1}}}, 
and find out for what value of {{{x}}} is  equal to {{{0}}}

{{{ x^2}}} – {{{4x}}} + {{{1}}} ={{{ 0}}}


We can use square root formula:

{{{x(1,2)=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}
   
{{{x(1,2)=(-(-4) +- sqrt ((-4)^2 -4*1*1 )) / (2*1)}}}

{{{x(1,2)=(4+- sqrt (16 -4 )) / (2)}}}

{{{x(1,2)=(4+- sqrt (12 )) / (2)}}}   

{{{x(1,2)=(4+- sqrt (4*3 )) / (2)}}}

{{{x(1,2)=(4+- 2sqrt (3 )) / (2)}}}

{{{x(1)=(4+ 2sqrt (3 )) / (2)}}}    

{{{x(1,2)=2(2+ sqrt (3 )) / (2)}}}     eliminate {{{2}}} in both nominator and denominator

{{{x(1)=2+ sqrt (3 ) }}}    

Use same way to find {{{x2}}}

 {{{x(2)=2 - sqrt (3 ) }}}     

 the real zeros are: {{{0}}}, {{{2+ sqrt(3)}}}, {{{and}}} {{{2 - (sqrt(3))}}}

since {{{sqrt(3) = 1.73}}}, we can write

{{{2+ sqrt(3)}}}= {{{2 + 1.73= 3.73}}}  {{{and }}}  second zero

{{{2}}}- {{{sqrt(3)}}}= {{{2 - 1.73}}}= {{{0.27}}}  third zero

so, the real zeros are: {{{0}}}, {{{3.73}}}, {{{and }}}{{{0.27}}}