3.04  Factoring by Synthetic Division

Rational Root Theorem, Descartes Rule

College Algebra: One Step at a Time,  Pages 378 - 405:   #9, 15, 21, 25, 27, 30, 31, 42, 43, 53

Special Graphing Calculator Solution for TI 84 (or TI 83+)!!  See Page 397:  #42, 43, 53

Dr. Robert J. Rapalje

Seminole Community College

Sanford, FL  32773

p. 380.  #9.  Factor  by grouping.

Solution:     Group the first two terms, and factor out the .  Group the last two terms, and factor out a  . 

Now factor out the common factor :

Factor the difference of squares:

                                           .

p. 384.  # 15.    , given that       is a root.

Solution:  If  is a root, then , so  is a factor of the polynomial.  Therefore you must do synthetic division by the number  in order to find the other factor, which gives you the reduced equation.  First write down the coefficients of the polynomial, which are  and prepare to do synthetic division with .

Of  the resulting numbers   , the last number  is the remainder.  (Of course it is ZERO!  If   is a factor, then the remainder MUST be!!)  Also, the first three numbers  are the coefficients of the quotient.  The quotient always begins with an exponent that is one less than the highest power of the polynomial.  The quotient will therefore begin with .  The quotient is  , and the reduced equation that must now be solved is .  Fortunately, this factors:

Final Answer:  There are three roots: 

p. 384.  # 21.    , given that       is a root.

Solution:  If  is a root, then , so  is a factor of the polynomial.  Therefore you must do synthetic division by the number  in order to find the other factor, which gives you the reduced equation.  First write down the coefficients of the polynomial, which are  and prepare to do synthetic division with .

 Of  the resulting numbers   , the last number  is the remainder.  (Of course it is ZERO!  If   is a factor, then the remainder MUST be!!)  Also, the first three numbers  are the coefficients of the quotient.  The quotient always begins with an exponent that is one less than the highest power of the polynomial.  The quotient will therefore begin with .  The quotient is  , and the reduced equation that must now be solved is .  Unfortunately, this does NOT factor, so you will have to solve it by quadratic formula or completing the square.  Completing the square in this case is actually easier.

Final Answer:       There are three roots: 

p. 386.  # 25.    , given that    are roots.

Solution:  If  is a root, then you must do synthetic division by the number  in order to find the other factor, which gives you a reduced equation.   As always, first write down the coefficients of the polynomial, which are  and prepare to do synthetic division with .

Of  the resulting numbers   ,  of course the last number  is the remainder, and the first four numbers  are the coefficients of the quotient.  Since the quotient always begins with an exponent that is one less than the highest power of the polynomial, the reduced equation begins with .   The reduced equation that must now be solved is .  This requires synthetic division this time b by .

Now the reduced equation is

which does NOT factor!  It must be solved by quadratic formula, completing the square, or graphing calculator methods.  Completing the square is a good way in this situation.

Take the square root of each side:

                   (also  and  from the synthetic division!)

Final Answer:  There are four roots:  .

p. 386.  # 27.    , given that       is a root.

Solution:  If  is a root, then you must do synthetic division by the number  in order to find the other factor, which gives you the reduced equation.  First write down the coefficients of the polynomial, which are  and prepare to do synthetic division with .

 Of  the resulting numbers   , the last number  is the remainder.  (Of course it is ZERO!  It  MUST be!!)  Also, the first three numbers  are the coefficients of the quotient.  The quotient always begins with an exponent that is one less than the highest power of the polynomial.  The quotient is  , and the reduced equation that must now be solved is , which can be solved by factoring:

                                  (also  that was given!)

Final Answer:  There are three roots: 

p. 388.  # 30.    , given that  there is a double root at  .

Solution:  If  is a root, then you must do synthetic division by the number  in order to find the other factor, which gives you a reduced equation.  The fact that it is a double root means that you will be able to use synthetic division with  twice.  As always, first write down the coefficients of the polynomial, which are  and prepare to do synthetic division (repeatedly) with .

Of  the resulting numbers   ,  of course the last number  is the remainder, and the first four numbers  are the coefficients of the quotient.  Since the quotient always begins with an exponent that is one less than the highest power of the polynomial, the reduced equation begins with .   The reduced equation that must now be solved is .  This requires synthetic division again by .

Now the reduced equation is

which factors into

                   (also  and  from the synthetic division!)

Final Answer:  There are four roots:  .

p. 388.  # 31.    , given that    is a multiple root.

Solution:  If    is a root, then you must do synthetic division by the number  in order to find the other factor, which gives you a reduced equation.  The fact that it is a multiple root means that you will be able to use synthetic division with  more than once.  As always, first write down the coefficients of the polynomial, which are  and prepare to do synthetic division (repeatedly) with .

Of  the resulting numbers   ,  of course the last number  is the remainder, and the first four numbers  are the coefficients of the quotient.  Since the quotient always begins with an exponent that is one less than the highest power of the polynomial, the reduced equation begins with .   The reduced equation that must now be solved is .  This requires synthetic division again by .

Now the reduced equation is

which factors into

                                      (also  and  from the synthetic division!)

Final Answer:  There are four roots: 

p. 393.  # 42.  Find all roots of  . 

Solution:  In Descartes’ day (or in my day for that matter!), it was necessary to find roots of a polynomial equation like this by trial and error process., using factors of 6 and synthetic division .  Today, we can find roots of a polynomial equation with a graphing calculator by either graphing techniques or a program in the TI 83+ or TI 84 called POLYSMLT.  (In the TI 85/86 look for [2^nd]  [POLY].  It works the same way!).  In the TI 83+ or TI 84, begin with the [APPS]  button, and in this menu, look for the program POLYSMLT, and press [ENTER].  The calculator asks for the Degree of the Polynomial (for a TI 85/86, the word “Order” is used instead of “Degree”—it means the same thing!), which is the highest power of x in the equation, which in this case is 3.  Type the number 3, and press [ENTER].  Next, enter each coefficient followed by [ENTER] beginning with the highest power.  Don’t forget, that if a term is missing, you must enter the coefficient as 0.  When you have entered all the coefficients, press the [F5] key, which represents [SOLVE].  The TI 84 comes back with this screen:

The roots (or zeros!) are obviously .  You can now do synthetic division with any of these three numbers, which will reduce the polynomial equation to a quadratic equation that can be solved by ordinary methods.  

As always, first write down the coefficients of the polynomial, which are  and prepare to do synthetic division (choose any of the  three numbers!!) with .

Of  the resulting numbers   ,  of course the last number  is the remainder, and the first three numbers  are the coefficients of the quotient.  Since the quotient always begins with an exponent that is one less than the highest power of the polynomial, the reduced equation begins with .   The reduced equation is

                     ,  which can be solved by factoring!

Final Answer:  There are three roots:    

ALTERNATE METHOD (especially for those who do NOT have POLYSMLT!!):

These roots can also be obtained by graphing , which gives this graph in the standard window. 

From this graph, you can see that the graph crosses the x axis at x = 1, -2, and -3.  With this information, you can use synthetic division with any of these roots, which reduces the equation to a quadratic equation to find the other two roots.  

p. 393.  # 43.  Find all roots of  . 

Solution:  In the past, it was necessary to find roots of a polynomial equation like this by trial and error process., using factors of 12 and synthetic division .  Today, we can find roots of a polynomial equation with a graphing calculator by either graphing techniques or a program in the TI 83+ or TI 84 called POLYSMLT.  (In the TI 85/86 look for [2^nd]  [POLY].  It works the same way!).  In the TI 83+ or TI 84, begin with the [APPS]  button, and in this menu, look for the program POLYSMLT, and press [ENTER].  The calculator asks for the Degree (or Order!) of the Polynomial, which is the highest power of x in the equation, which in this case is 3.  Type the number 3, and press [ENTER].  Next, enter each coefficient followed by [ENTER] beginning with the highest power.  Don’t forget, that if a term is missing, you must enter the coefficient as  0.  When you have entered all the coefficients, press the [F5] key, which represents [SOLVE].  The TI 84 comes back with this screen:

The roots (or zeros!) are obviously .  You can now do synthetic division with any of these three numbers, which will reduce the polynomial equation to a quadratic equation that can be solved by ordinary methods.  

As always, first write down the coefficients of the polynomial, which are  and prepare to do synthetic division (choose any of the  three numbers!!) with .

Of  the resulting numbers   ,  of course the last number  is the remainder, and the first three numbers  are the coefficients of the quotient.  Since the quotient always begins with an exponent that is one less than the highest power of the polynomial, the reduced equation begins with .   The reduced equation is

                     ,  which can be solved by factoring!

Final Answer:  There are three roots:  . 

ALTERNATE METHOD (especially for those who do NOT have POLYSMLT!!):

These roots can also be obtained by graphing , which gives this graph in the standard window.  The graph below, obtained by using [TRACE] with x=2,  illustrates the root or zero at x=2 .  Of course the zeros in this case can be easily seen from just looking at the graph, or from [2^nd]  [F5 (TABLE)]. 

From this graph, you can see that the graph crosses the x axis at x = 2, -2, and -3.  With this information, you can use synthetic division with any of these roots, which reduces the equation to a quadratic equation to find the other two roots.  

p. 397.  # 53.  Find all roots of  . 

Solution:  In Descartes’ day (or in my day for that matter!), it was necessary to find roots of a polynomial equation like this by trial and error process., using factors of 12 and synthetic division .  Today, we can find roots of a polynomial equation with a graphing calculator by either graphing techniques or a program in the TI 83+ or TI 84 called POLYSMLT.  (In the TI 85/86 look for [2^nd]  [POLY].  It works the same way!).  In the TI 83+ or TI 84, begin with the [APPS]  button, and in this menu, look for the program POLYSMLT, and press [ENTER].  The calculator asks for the Degree of the Polynomial, which is the highest power of x in the equation, which in this case is 5.  Type the number 5, and press [ENTER].  Next, enter each coefficient followed by [ENTER] beginning with the highest power.  Don’t forget, that if a term is missing, you must enter the coefficient as 0.  When you have entered all the coefficients, press the [F5] key, which represents [SOLVE].  It may take a few seconds to solve, but the TI 84 comes back with this screen:

It looks as if the calculator is giving you roots at .  The middle numbers, however, are misleading.  If you scroll down to one of these numbers and press the right arrow, you will see that these are actually   The E -7 is the scientific notation for a number divided by 10 million, which is essentially zero!  This means that these two middle numbers actually represent a double root at .  You can now do synthetic division consecutively with the three numbers which will reduce the fifth degree polynomial equation down to a quadratic equation (degree 2) that can be solved by ordinary methods.   

As always, first write down the coefficients of the polynomial, which are  and prepare to do synthetic division (repeatedly) beginning with .

Of  the resulting numbers   ,  of course the last number  is the remainder, and the first five numbers  are the coefficients of the quotient.  Since the quotient always begins with an exponent that is one less than the highest power of the polynomial, the reduced equation begins with .   The reduced equation that must now be solved is .  Next, perform synthetic division by .

This time the reduced equation is

Perform synthetic division by  again, since it is a double root (it occurs twice!).

Now the reduced equation is        or  .

This can be solved by adding -1 to each side of the equation:

and then take the square root of each side:

                                                        (also  and  from the synthetic division!)

Final Answer:  There are five roots: 

ALTERNATE METHOD (especially for those who do NOT have or do not like POLYSMLT!!):

These roots can also be obtained by graphing , which gives this graph in the standard window. 

Adjusting the window may make it easier to see what the graph really does. 

I suggest using a window such as x=[-5,5] and y=[-100,100]:

From either of these graphs, even without knowing what the graph actually looks like, you can see that the graph crosses the x axis at x = -3, and it bounces at x = 2.  This means that there is an even multiplicity (i.e., a double root!) at x = 2.  With this information, you can proceed to use synthetic division to find the other two roots which happen to be imaginary roots.