# SOLUTION: Use synthetic division grid to analyze and graph the function. P(x) = 9x^4 – 6x^3 – 26x^2 + 18 x - 3

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Use synthetic division grid to analyze and graph the function. P(x) = 9x^4 – 6x^3 – 26x^2 + 18 x - 3       Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Exponent and logarithm as functions of power Solvers Lessons Answers archive Quiz In Depth

Question 135713: Use synthetic division grid to analyze and graph the function.
P(x) = 9x^4 – 6x^3 – 26x^2 + 18 x - 3

Answer by jim_thompson5910(28715)   (Show Source):
You can put this solution on YOUR website!
Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients

So let's list the factors of 3 (the last coefficient):

Now let's list the factors of 9 (the first coefficient):

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

Now simplify

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

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 8 possible roots, so that means there would be at most 8 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

It turns out that is actually a root. So our test zero is 1/3

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.
 1/3 | 9 -6 -26 18 -3 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 9)
 1/3 | 9 -6 -26 18 -3 | 9

Multiply 1/3 by 9 and place the product (which is 3) right underneath the second coefficient (which is -6)
 1/3 | 9 -6 -26 18 -3 | 3 9

Add 3 and -6 to get -3. Place the sum right underneath 3.
 1/3 | 9 -6 -26 18 -3 | 3 9 -3

Multiply 1/3 by -3 and place the product (which is -1) right underneath the third coefficient (which is -26)
 1/3 | 9 -6 -26 18 -3 | 3 -1 9 -3

Add -1 and -26 to get -27. Place the sum right underneath -1.
 1/3 | 9 -6 -26 18 -3 | 3 -1 9 -3 -27

Multiply 1/3 by -27 and place the product (which is -9) right underneath the fourth coefficient (which is 18)
 1/3 | 9 -6 -26 18 -3 | 3 -1 -9 9 -3 -27

Add -9 and 18 to get 9. Place the sum right underneath -9.
 1/3 | 9 -6 -26 18 -3 | 3 -1 -9 9 -3 -27 9

Multiply 1/3 by 9 and place the product (which is 3) right underneath the fifth coefficient (which is -3)
 1/3 | 9 -6 -26 18 -3 | 3 -1 -9 3 9 -3 -27 9

Add 3 and -3 to get 0. Place the sum right underneath 3.
 1/3 | 9 -6 -26 18 -3 | 3 -1 -9 3 9 -3 -27 9 0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 4 coefficients (9,-3,-27,9) form the quotient

Notice in the denominator , the x term has a coefficient of 3, so we need to divide the quotient by 3 like this:

So

You can use this online polynomial division calculator to check your work

Basically factors to

Now lets break down further

Once again through the rational root theorem, we find that is another root (since it occurs twice)

So once again our test zero is 1/3

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.
 1/3 | 3 -1 -9 3 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 3)
 1/3 | 3 -1 -9 3 | 3

Multiply 1/3 by 3 and place the product (which is 1) right underneath the second coefficient (which is -1)
 1/3 | 3 -1 -9 3 | 1 3

Add 1 and -1 to get 0. Place the sum right underneath 1.
 1/3 | 3 -1 -9 3 | 1 3 0

Multiply 1/3 by 0 and place the product (which is 0) right underneath the third coefficient (which is -9)
 1/3 | 3 -1 -9 3 | 1 0 3 0

Add 0 and -9 to get -9. Place the sum right underneath 0.
 1/3 | 3 -1 -9 3 | 1 0 3 0 -9

Multiply 1/3 by -9 and place the product (which is -3) right underneath the fourth coefficient (which is 3)
 1/3 | 3 -1 -9 3 | 1 0 -3 3 0 -9

Add -3 and 3 to get 0. Place the sum right underneath -3.
 1/3 | 3 -1 -9 3 | 1 0 -3 3 0 -9 0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (3,0,-9) form the quotient

Notice in the denominator , the x term has a coefficient of 3, so we need to divide the quotient by 3 like this:

So

You can use this online polynomial division calculator to check your work

So this means that factors to

This tells us that the zeros are , , and (with a multiplicity of 2)

So we can now graph the function by drawing a curve through the zeros