Question 970518
Given 

{{{h(x)=2x^3- 5x^2 -4x + 3}}}

 Find the intercepts of {{{h(x)}}}, or find the zeros

{{{2x^3- 5x^2 -4x + 3=0}}}

{{{2x^3- 6x^2+x^2 -3x -x+ 3=0}}}

{{{(2x^3- 6x^2)+(x^2 -3x) -(x- 3)=0}}}

{{{2x^2(x- 3)+x(x -3) -(x- 3)=0}}}

{{{(2x^2+x-1)(x-3)=0}}}

{{{(2x^2+2x-x-1)(x-3)=0}}}

{{{((2x^2+2x)-(x+1))(x-3)=0}}}

{{{(2x(x+1)-(x+1))(x-3)=0}}}

{{{(2x-1) (x+1) (x-3)=0}}}

solutions:

if {{{(2x-1) =0}}}=>{{{2x=1}}}=>{{{x=1/2}}}

if {{{ (x+1)=0}}}=>{{{x=-1}}}

if {{{(x-3)=0}}}=>{{{x=3}}}

so, the intercepts are at:

({{{1/2}}},{{{0}}})

({{{-1}}},{{{0}}})

({{{3}}},{{{0}}})



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