Question 90672
Any rational zero can be found through this equation


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



So let's list the factors of 12:


*[Tex \LARGE p=\pm1, \pm2, \pm3, \pm4, \pm6, \pm12]


Now let's list the factors of 1:


*[Tex \LARGE q=\pm1]


Now let's divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{2}{1}, \frac{3}{1}, \frac{4}{1}, \frac{6}{1}, \frac{12}{1}, \frac{-1}{1}, \frac{-2}{1}, \frac{-3}{1}, \frac{-4}{1}, \frac{-6}{1}, \frac{-12}{1}]


Now simplify


These are all the possible zeros of the function


*[Tex \LARGE 1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12]



To save time, I'm only going to use synthetic division on the possible zeros that are actually zeros of the function.
Otherwise, I would have to use synthetic division on every possible root (there are 12 possible roots, so that means there would be at most 12 synthetic division tables).
However, you might be required to follow this procedure, so this is why I'm showing you how to set up a problem like this



If you're not required to follow this procedure, simply use a graphing calculator to find the roots



So with a graphing calculator, we find a root x=-4. So our test zero is -4



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>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</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>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</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 -4 by 1 and place the product (which is -4)  right underneath the second  coefficient (which is 3)

    <TABLE cellpadding=10><TR><TD>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-4</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 -4 and 3 to get -1. Place the sum right underneath -4.

    <TABLE cellpadding=10><TR><TD>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-4</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -4 by -1 and place the product (which is 4)  right underneath the third  coefficient (which is -1)

    <TABLE cellpadding=10><TR><TD>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-4</TD><TD>4</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD></TR></TABLE>

    Add 4 and -1 to get 3. Place the sum right underneath 4.

    <TABLE cellpadding=10><TR><TD>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-4</TD><TD>4</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>3</TD><TD></TD></TR></TABLE>

    Multiply -4 by 3 and place the product (which is -12)  right underneath the fourth  coefficient (which is 12)

    <TABLE cellpadding=10><TR><TD>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-4</TD><TD>4</TD><TD>-12</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>3</TD><TD></TD></TR></TABLE>

    Add -12 and 12 to get 0. Place the sum right underneath -12.

    <TABLE cellpadding=10><TR><TD>-4</TD><TD>|</TD><TD>1</TD><TD>3</TD><TD>-1</TD><TD>12</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-4</TD><TD>4</TD><TD>-12</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>3</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+4}}} is a factor of  {{{x^3 + 3x^2 - x + 12}}}


Now lets look at the bottom row of coefficients:


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


{{{x^2 - x + 3}}}



So {{{(x^3 + 3x^2 - x + 12)/(x+4)=x^2 - x + 3}}}


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



Now let's use the quadratic formula to solve {{{x^2 - x + 3}}}



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-x+3=0}}} ( notice {{{a=1}}}, {{{b=-1}}}, and {{{c=3}}})


{{{x = (--1 +- sqrt( (-1)^2-4*1*3 ))/(2*1)}}} Plug in a=1, b=-1, and c=3




{{{x = (1 +- sqrt( (-1)^2-4*1*3 ))/(2*1)}}} Negate -1 to get 1




{{{x = (1 +- sqrt( 1-4*1*3 ))/(2*1)}}} Square -1 to get 1  (note: remember when you square -1, you must square the negative as well. This is because {{{(-1)^2=-1*-1=1}}}.)




{{{x = (1 +- sqrt( 1+-12 ))/(2*1)}}} Multiply {{{-4*3*1}}} to get {{{-12}}}




{{{x = (1 +- sqrt( -11 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (1 +- i*sqrt(11))/(2*1)}}} Simplify the square root (note: since we cannot take the square root of a negative value, we must factor {{{sqrt(-11)}}} to {{{i*sqrt(11)}}} to make the radicand positive. 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 = (1 +- i*sqrt(11))/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=1/2 + sqrt(11)/2*i}}} or {{{x=1/2 - sqrt(11)/2*i}}}


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So the polynomial {{{x^3 + 3x^2 - x + 12}}} has the roots



{{{x=-4}}}, {{{x=1/2 + sqrt(11)/2*i}}} or {{{x=1/2 - sqrt(11)/2*i}}}