Question 383381
{{{graph(300,300,-3,3,-3,3,0,x^3+2x^2-5x-10)}}}
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The real zeros lies between ({{{-3}}},{{{3}}}).
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{{{graph(300,300,-3,1,-1,1,0,x^3+2x^2-5x-10)}}}
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Looks like {{{x=-2}}} is a zero.
Verify using the equation,
{{{x^3+2x^2-5x-10=(-2)^3+2(-2)^2-5(-2)-10}}}
{{{x^3+2x^2-5x-10=(-8)+(8)+10-10}}}
{{{x^3+2x^2-5x-10=0}}}
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Since {{{x=-2}}} is a zero, then {{{x+2}}} is a factor.
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Use polynomial long division to find the other quadratic factor,

<pre>

        x^2 - 5
       __________________________
x + 2 | x^3 + 2x^2 - 5x - 10
     - (x^3 + 2x^2)
       ---------------------
                   - 5x - 10
                - (- 5x - 10)
                -------------
                           0
</pre>
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{{{(x^3 + 2x^2 - 5x - 10)=(x+2)(x^2-5)}}}
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For the quadratic equation,
{{{x^2-5=0}}}
{{{x^2=5}}}
{{{x=0 +- sqrt(5)}}}
You can approximate those zeros with a calculator to the desired number of decimal places.