Algebra Factoring Higher Degree Polynomials Synthetic Division Synthetic Division is a short‐cut to dividing polynomials by a linear factor. Here’s how it works. We will use an example to illustrate the process. Example 1: 2
2
5
2 Step 1: In the linear term
take the value r as the divisor. In the example, the divisor will be . We use the letter r to indicate that the value is actually a root of the equation. So, in synthetic division, the root is used as the divisor. Step 2: Line up the coefficients of the terms from highest degree to lowest degree in a row to the right of the divisor. If a term is missing, use a zero for the coefficient of that term. We will call this array of coefficients the dividend. 2
‐2
5
1
‐2
5
1
‐2
1
‐2
‐4 ‐2
2
1 ‐1
0
2
Step 3: Bring the leading coefficient down below the line. Step 4: Multiply the divisor by the number just placed below the line and put the result above the line and one column to the right. Add the two numbers in that column to get a number below the line for that column. ‐2
Step 5: Repeat Step 4 until all of the columns have been completed. ‐2
2
‐4
The final result is a set of coefficients of the polynomial that results from the division. The exponents of the terms of the resulting polynomial begin one lower than the degree of the original polynomial. 2
1
2
5
2
1 rem
In the example, the result is , with a remainder of 0. The remainder of 0 is a good indication that the division was performed properly. Example 2: 3
4
1 1
From the synthetic division to the right, we get: There is no constant term and no remainder in the solution to this example. Version 2.7
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0
3
0
‐4
0
1
1
4
4
0
1
4
4
0
0
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Algebra Comparing Synthetic Division to Long Division Advantages of Synthetic Division Synthetic division has the following advantages over long division: •
•
•
•
The divisor is a possible root of the polynomial; it is a root if the remainder is zero. It is shorter. It is much quicker. It works by addition and multiplication instead of by subtraction and division. Because of this, it is much less prone to error. Comparison of Methods It is instructive to compare synthetic division and long division to get a better idea of why synthetic division works. Consider the division: 2
5
2
2 The two methods of performing this division are laid out below. Notice the following correspondences between the examples: •
•
•
•
•
Synthetic Division Root vs. Factor. Synthetic division uses the root of the polynomial as the divisor. Long division uses the whole factor. The signs on the root are opposite in the two methods. ‐2
2
Dividend. The dividends in the two methods are the same (except that synthetic division leaves out the variables). Second Row. The second row in synthetic division corresponds to the “secondary” coefficients of each division in long division (but with opposite signs). Answer Row. In synthetic division the answer row (of coefficients) is calculated directly by adding the values in the rows above it. In long division, it is necessary to subtract expressions to determine another expression that must be divided by the divisor to get the next term of the answer. Adding Variables. In synthetic division, it is necessary to add the variables after the answer is determined. In long division, the answer is provided directly. 5
2
1
‐2
‐4 ‐2
2
1 ‐1
0
Long Division 2
1
1 2 2
2
1
2 5
4
2 1
2 1
1
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Algebra Zeroes of Polynomials Developing Possible Roots If a polynomial has the form: Then, •
will have exactly complex roots. For example, a 5th degree polynomial will have exactly 5 complex roots. Note: some of these roots may be the same, and some of them may be real. •
will have exactly real roots, where is a whole number. For example, a 5 degree polynomial will have either 5 real roots, 3 real roots, or 1 real root. th
•
Descartes’ Rule of Signs. (Note how this ties into the bullet above.) o The number of positive real roots of a polynomial is equal to the number of sign changes in , or is less than this by a multiple of 2. o The number of negative real roots of a polynomial is equal to the number of sign changes in , or is less than this by a multiple of 2. Note: to generate quickly, just change the signs of the terms with odd exponents. •
will have an even number of non‐real roots. For example, a 5th degree polynomial will have either 0 non‐real roots, 2 non‐real roots, or 4 non‐real roots. Further, the non‐real roots exist in conjugate pairs; so if is a root of , then so is . •
Rational Root Theorem. Any rational roots have the characteristic . This fact is especially useful if the lead coefficient of the polynomial is 1; in this case, any real roots are factors of the constant term. This fact, in combination with the ease of synthetic division, makes finding integer roots a quick process. Example: What can we say about the roots of ? (note: 4 sign changes) •
First, note that •
So, •
The real roots must be 1, 2, or 4 (the positive factors of the constant term 4). •
To find out more, we have to test the possible real root values. Version 2.7
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Algebra Zeroes of Polynomials Testing Possible Roots The following two theorems are very useful in testing possible roots (zeroes) of Polynomials. Factor Theorem: is a factor of a polynomial Remainder Theorem: If is divided by 0. if and only if . , then the remainder is Methods of Testing Possible Roots If a polynomial can be factored, then first, factor the polynomial; the problem will be easier to solve after factoring. In addition, if you are able to produce linear or quadratic factors, the roots of those factors will be roots of the polynomial. After factoring, the following methods can be used to test possible roots of a polynomial. •
Use synthetic division to test possible roots. Because synthetic division is quick, several potential roots can be tested in a short period of time. •
Substitute possible roots into the polynomial to see if the remainder is zero. If , then is a root of . •
Graph the polynomial. Real roots exist wherever the graph crosses the x‐axis. Although this method may help find the approximate location of roots, it is not a reliable method for determining exact values of roots. Example: Factor and find the roots of Using synthetic division: 1
1
‐2
1
2
1
‐4
1 ‐1
0 ‐4
‐1
‐4
0
1
1
Trying first the possible root 1, then the possible root 2, we find that they both work. So, 4
0
‐1
0 ‐4
2
2
4
1
2
0
Using the quadratic formula on the quadratic factor in this expression we find two non‐real roots. So the four roots are: , ,
√
,
√
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