Factoring & Synthetic Division

Factoring with Synthetic Division

Method

1) By examining the factors of the numerical coefficient of the highest degree term and the constant term you can form test ratios that may be roots of the polynomial.

2) These ratios can be tested by performing synthetic division on the numerator. If the final result is zero, the test ratio is a root of the polynomial, and the root can be used to form one factor while the other coefficients in the result of the synthetic division form another polynomial factor.

3) If the polynomial is in an equation set equal to zero, those test ratios indicate the solutions or roots of the equation.

Study the Examples in this subsection to see this method illustrated.

Example 1:

Factor and solve the following 4th degree polynomial.

I will use synthetic division to help factor the polynomial. Note below how the coefficients of the polynomial appear in the first row of the form for the synthetic division. We need to find a divisor like ( x + 2 ), which is represented by the - 2 in the form, that will produce a zero remainder in the last column of the third row of the form.

To guess a root of the polynomial, like the - 2 below, try a factor of the constant term in the polynomial : 12. These factors are : .

Here is what you get using the - 2 as a possible root :

Your guess is confirmed only if you get a zero in the final position. When that happens the last row of the form indicates one of the factors, in this case it is a cubic. The corresponding linear expression (x+2) is also confirmed as a factor. If you do not get a 0 remainder in the last position of the synthetic division, you will have to try another possibility.

We now can take the cubic factor from the results of the synthetic division above to repeat the process of looking for another root from this cubic and to test it with synthetic division. Note that for possible whole number roots of this cubic we need only to look at a shorter list of factors of the constant term : 6. These factors are : .

Lets try a root of 2 whose factor would be (x-2).

That worked too. It must be our lucky day! Since the quadratic can not be factored with whole numbers we leave it as is. Thus, the final factored form of the polynomial is :

You can get the last two roots using the quadratic formula. See the section on quadratic equations for a review of the use of this formula. The four roots of this polynomial are :

Important Rule : The only possible whole number roots of a polynomial must be a factor of the constant term of the polynomial.

Example 2:

Factor the following cubic ploynomial : .

We set up the first row of the synthetic division form with the polynomials coefficients : 2, 3, -8, and 3.

When we try ( -1) as a possible root we get a remainder of 11, not zero :

That doesn't look good, so lets try (+1).

That worked ( we got a zero remainder ), thus is a factor of and is the other quadratic factor as seen from the last row of the division.

This quadratic factors into so the complete factorization of the polynomial is :

Example 3:

Lets factor and solve this cubic polynomial. This one is a little harder to factor, because the cubic term has a coefficient of 2.

The possible roots are listed below. Do you see how I found them?

Many of these will fail but if we try 1/2 we get the desired zero remainder.

This confirms (x - 1/2) as a factor. Can you write the other quadraic factor from the third row above? Try to factor this further and click on the button below to see those factors.

Now the completely factored expression. See if you can create it before clicking the button.

Now the solutions or roots. Try again to write them down before clicking the button.

Important Rule : Possible fractional roots of a polynomial must use a factor from the constant term of the polynomial in the numerator and a factor from the leading coefficient of the polynomial in the denominator of the root.

Example 4:

Solve the following cubic equation. The following example has a missing linear term. This causes a minor problem in factoring.

Now that you are getting fairly good at starting this process, write down the form and a root for factoring the cube.

Do you know why there is a 0 in the setup? We have to use 0 as a place holder for the missing linear term. So what does the synthetic division tell you? Factor the quadratic further and click the button to see the final factors.