Question 138661
First graph the function


{{{ graph( 500, 500, -10, 10, -500, 500,x^4-2x^3-19x^2+122x-312) }}} Graph of {{{y=x^4-2x^3-19x^2+122x-312}}}



From the graph we can see that the function has a zero at {{{x=-6}}}. So our test zero is -6. So our first factor is {{{x+6}}}




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>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><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><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>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><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><TD></TD></TR></TABLE>

    Multiply -6 by 1 and place the product (which is -6)  right underneath the second  coefficient (which is -2)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -6 and -2 to get -8. Place the sum right underneath -6.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -6 by -8 and place the product (which is 48)  right underneath the third  coefficient (which is -19)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>48</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 48 and -19 to get 29. Place the sum right underneath 48.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>48</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -6 by 29 and place the product (which is -174)  right underneath the fourth  coefficient (which is 122)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>48</TD><TD>-174</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD></TD><TD></TD></TR></TABLE>

    Add -174 and 122 to get -52. Place the sum right underneath -174.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>48</TD><TD>-174</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD><TD></TD></TR></TABLE>

    Multiply -6 by -52 and place the product (which is 312)  right underneath the fifth  coefficient (which is -312)

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>48</TD><TD>-174</TD><TD>312</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD><TD></TD></TR></TABLE>

    Add 312 and -312 to get 0. Place the sum right underneath 312.

    <TABLE cellpadding=10><TR><TD>-6</TD><TD>|</TD><TD>1</TD><TD>-2</TD><TD>-19</TD><TD>122</TD><TD>-312</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-6</TD><TD>48</TD><TD>-174</TD><TD>312</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x+6}}} is a factor of  {{{x^4 - 2x^3 - 19x^2 + 122x - 312}}}


Now lets look at the bottom row of coefficients:


The first 4 coefficients (1,-8,29,-52) form the quotient


{{{x^3 - 8x^2 + 29x - 52}}}



So {{{(x^4 - 2x^3 - 19x^2 + 122x - 312)/(x+6)=x^3 - 8x^2 + 29x - 52}}}


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


Basically  {{{x^4 - 2x^3 - 19x^2 + 122x - 312}}} factors to {{{(x+6)(x^3 - 8x^2 + 29x - 52)}}}


Now lets break  {{{x^3 - 8x^2 + 29x - 52}}} down further



Also from the graph, we can see that the function has another zero at {{{x=4}}}. So our test zero is 4. So our second factor is {{{x-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>-8</TD><TD>29</TD><TD>-52</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>-8</TD><TD>29</TD><TD>-52</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 -8)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</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 -8 to get -4. Place the sum right underneath 4.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</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>-4</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 4 by -4 and place the product (which is -16)  right underneath the third  coefficient (which is 29)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-16</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-4</TD><TD></TD><TD></TD></TR></TABLE>

    Add -16 and 29 to get 13. Place the sum right underneath -16.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-16</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-4</TD><TD>13</TD><TD></TD></TR></TABLE>

    Multiply 4 by 13 and place the product (which is 52)  right underneath the fourth  coefficient (which is -52)

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-16</TD><TD>52</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-4</TD><TD>13</TD><TD></TD></TR></TABLE>

    Add 52 and -52 to get 0. Place the sum right underneath 52.

    <TABLE cellpadding=10><TR><TD>4</TD><TD>|</TD><TD>1</TD><TD>-8</TD><TD>29</TD><TD>-52</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-16</TD><TD>52</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-4</TD><TD>13</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 - 8x^2 + 29x - 52}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,-4,13) form the quotient


{{{x^2 - 4x + 13}}}



So {{{(x^3 - 8x^2 + 29x - 52)/(x-4)=x^2 - 4x + 13}}}


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 - 8x^2 + 29x - 52}}} factors to {{{(x-4)(x^2 - 4x + 13)}}}


Now lets break  {{{x^2 - 4x + 13}}} 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^2-4*x+13=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=13}}})





{{{x = (--4 +- sqrt( (-4)^2-4*1*13 ))/(2*1)}}} Plug in a=1, b=-4, and c=13




{{{x = (4 +- sqrt( (-4)^2-4*1*13 ))/(2*1)}}} Negate -4 to get 4




{{{x = (4 +- sqrt( 16-4*1*13 ))/(2*1)}}} Square -4 to get 16  (note: remember when you square -4, you must square the negative as well. This is because {{{(-4)^2=-4*-4=16}}}.)




{{{x = (4 +- sqrt( 16+-52 ))/(2*1)}}} Multiply {{{-4*13*1}}} to get {{{-52}}}




{{{x = (4 +- sqrt( -36 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (4 +- 6*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 = (4 +- 6*i)/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=2 + 3*i}}} or {{{x=2 - 3*i}}}





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Answer:


So the zeros of the original polynomial {{{y=x^4-2x^3-19x^2+122x-312}}} are:



{{{x=-6}}}, {{{x=4}}}, {{{x=2 + 3*i}}} and {{{x=2 - 3*i}}}