Question 92193
If -2 is a root, then -2 is a test zero. So that means we can use synthetic division


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.<TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

<TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -2 by 1 and place the product (which is -2)  right underneath the second  coefficient (which is 2)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -2 and 2 to get 0. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -2 by 0 and place the product (which is 0)  right underneath the third  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>0</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD></TD><TD></TD></TR></TABLE>

    Add 0 and 1 to get 1. Place the sum right underneath 0.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>0</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>1</TD><TD></TD></TR></TABLE>

    Multiply -2 by 1 and place the product (which is -2)  right underneath the fourth  coefficient (which is 2)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>0</TD><TD>-2</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>1</TD><TD></TD></TR></TABLE>

    Add -2 and 2 to get 0. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>0</TD><TD>-2</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>1</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+2}}} is a factor of  {{{x^3 + 2x^2 + x + 2}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,0,1) form the quotient


{{{x^2 + 1}}}



So {{{(x^3 + 2x^2 + x + 2)/(x+2)=x^2 + 1}}}


You can use this <a href=http://calc101.com/webMathematica/long-divide.jsp>online polynomial division calculator</a> to check your work




Basically  {{{x^3 + 2x^2 + x + 2}}} factors to {{{(x+2)(x^2 + 1)}}}


Now lets break  {{{x^2 + 1}}} down further




Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{x^2+1=0}}} (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like {{{x^2+0*x+1=0}}}  notice {{{a=1}}}, {{{b=0}}}, and {{{c=1}}})


{{{x = (0 +- sqrt( (0)^2-4*1*1 ))/(2*1)}}} Plug in a=1, b=0, and c=1




{{{x = (0 +- sqrt( 0-4*1*1 ))/(2*1)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+-4 ))/(2*1)}}} Multiply {{{-4*1*1}}} to get {{{-4}}}




{{{x = (0 +- sqrt( -4 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (0 +- 2*i)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (0 +- 2*i)/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=0 + i}}} or {{{x=0 - i}}}



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Answer:


So the polynomial has roots 


{{{x=-2}}}, {{{x=i}}} and {{{x=-i}}} (the last two are the imaginary roots)