Question 320120
{{{ graph( 300, 300, -5, 5, -5, 5,  x^3 + 3x^2 + 4x-8)}}}
Looks like {{{x=1}}} is a zero.
Verify using the equation.
{{{1+3+4-8=0}}}
Yes, so {{{x-1}}} is a factor.
Use polynomial long division to find the remaining quadratic factor.
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First term: {{{x^2}}}
{{{x^2(x-1)=x^3-x^2}}}
Subtract from the original polynomial to get the remainder,
{{{(x^3+3x^2+4x-8)-(x^3-x^2)=4x^2+4x-8}}}
Next term: {{{4x}}}
{{{4x(x-1)=4x^2-4x}}}
Subtract from the first remainder to get the next remainder,
{{{(4x^2+4x-8)-(4x^2-4x)=8x-8}}}
Final term:{{{8}}}
{{{8(x-1)=8x-8}}}
Subtract from the second remainder to get the next remainder,
{{{(8x-8)-(8x-8)=0}}}
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{{{(x^3+3x^2+4x-8)/(x-1)=x^2+4x+8}}}
Solve for the remaining zeros by completing the square.
{{{x^2+4x+8=0}}}
{{{x^2+4x+4+8-4=0}}}
{{{(x+2)^2+4=0}}}
{{{(x+2)^2=-4}}}
{{{x+2=0 +- 2i}}}
{{{highlight(x=-2 +- 2i)}}}
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{{{x=1}}}
{{{x=-2+2i}}}
{{{x=-2-2i}}}