Question 148299
i)


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 -3 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm3]


Now let's list the factors of 2 (the first coefficient):


*[Tex \LARGE q=\pm1, \pm2]


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



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







Now simplify


These are all the distinct rational zeros of the function that could occur


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






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ii)


With the help of a graphing calculator, we see that -1 is a zero of {{{2x^3-3x^2-8x-3}}}


note: let me know if you need to find the zeros a different way.


So let's set up a synthetic division table by placing the value -1 in the upper left corner and placing the coefficients of the polynomial to the right of -1.<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</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 2)

<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by 2 and place the product (which is -2)  right underneath the second  coefficient (which is -3)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</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>2</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -2 and -3 to get -5. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</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>2</TD><TD>-5</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1 by -5 and place the product (which is 5)  right underneath the third  coefficient (which is -8)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>5</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-5</TD><TD></TD><TD></TD></TR></TABLE>

    Add 5 and -8 to get -3. Place the sum right underneath 5.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>5</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-5</TD><TD>-3</TD><TD></TD></TR></TABLE>

    Multiply -1 by -3 and place the product (which is 3)  right underneath the fourth  coefficient (which is -3)

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>5</TD><TD>3</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-5</TD><TD>-3</TD><TD></TD></TR></TABLE>

    Add 3 and -3 to get 0. Place the sum right underneath 3.

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>2</TD><TD>-3</TD><TD>-8</TD><TD>-3</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>5</TD><TD>3</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-5</TD><TD>-3</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, this means that -1 is a zero of {{{2x^3-3x^2-8x-3}}} (this confirms our original claim). 




Now lets look at the bottom row of coefficients:


The first 3 coefficients (2,-5,-3) form the quotient


{{{2x^2 - 5x - 3}}}



So {{{(2x^3 - 3x^2 - 8x - 3)/(x+1)=2x^2 - 5x - 3}}}



Basically  {{{2x^3 - 3x^2 - 8x - 3}}} factors to {{{(x+1)(2x^2 - 5x - 3)}}}



Now lets find the zeros for  {{{2x^2 - 5x - 3}}}.



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-5) +- sqrt( (-5)^2-4(2)(-3) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-5}}}, and {{{c=-3}}}



{{{x = (5 +- sqrt( (-5)^2-4(2)(-3) ))/(2(2))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{x = (5 +- sqrt( 25-4(2)(-3) ))/(2(2))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25--24 ))/(2(2))}}} Multiply {{{4(2)(-3)}}} to get {{{-24}}}



{{{x = (5 +- sqrt( 25+24 ))/(2(2))}}} Rewrite {{{sqrt(25--24)}}} as {{{sqrt(25+24)}}}



{{{x = (5 +- sqrt( 49 ))/(2(2))}}} Add {{{25}}} to {{{24}}} to get {{{49}}}



{{{x = (5 +- sqrt( 49 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (5 +- 7)/(4)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (5 + 7)/(4)}}} or {{{x = (5 - 7)/(4)}}} Break up the expression. 



{{{x = (12)/(4)}}} or {{{x =  (-2)/(4)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -1/2}}} Simplify. 



So the zeros of {{{2x^3-3x^2-8x-3}}} are {{{x=-1}}}, {{{x=3}}}, or {{{x=-1/2}}}