12 Synthetic Division by x − aIN ARITHMETIC we write, for example, or,
Equivalently, 47 = 9· 5 + 2 5 is called the divisor, 47 is the dividend, 9 is the quotient, and 2 is the remainder.
Or, Dividend = Quotient· Divisor + Remainder. In algebra, if we divide a polynomial P(x) by a polynomial D(x) (where the degree of D is less than the degree of P), we would find P(x) = Q(x)· D(x) + R(x). P(x) is the dividend, Q(x) is the quotient, and R(x) is the remainder. For example, if, by long division, we divided x^{3} − 5x^{2} + 3 x − 7 by x − 2, we would find
Or, = (x^{2} − 3x − 3)(x − 2) − 13. x^{3} − 5x^{2} + 3x − 7 is the dividend, x^{2} − 3x −3 is the quotient, and −13 is the remainder. Here is how to do this problem by synthetic division. First, to use synthetic division, the divisor must be of the first degree and must have the form x − a. In this example, the divisor is x − 2, with a = 2. Here again is the problem:
Proceed as follows: 1. Write the coefficients of the dividend: 1 − 5 + 3 − 7 2. Put a, in this case 2, in a box to the right, leave a space, and draw a 3. Bring down the leading coefficient (1), multiply it with a (2), and
4. Add: 5. Repeat the process. −3· 2 = −6. And so on, until all the coefficients
The first three numbers, 1 − 3 − 3, are the coefficients of the quotient, and the final number, −13, is the remainder. We have x^{3} − 5x^{2} + 3 x − 7 = (x^{2} − 3x −3)(x − 2) − 13. Example 1. Use synthetic division to divide 2x^{5} + 3x^{4} + 25x^{2} − 1 by x + 3. Solution. There are a couple of points here. First, we must account for all six coefficients of the general form. 2 + 3 + 0 + 25 + 0 − 1 The coefficient of x^{3} is 0, as is the coefficient of x. Next, the divisor is x + 3. But the divisor must have the form x − a. x + 3 = x − (−3). Therefore, a = −3. Here is the synthetic division: This tells us
Or,
Note: The degree of the quotient is one less than the degree of the dividend. And the degree of the remainder is less than the degree of the divisor, x + 3, which in this case is 1. The remainder therefore is of degree 0, which is a number. In general, if we divide a polynomial of degree n by a polynomial of degree 1, then the degree of the quotient will be n − 1. And the remainder will be a number. Problem 1. Use synthetic division to divide x^{3} − 8x^{2} + x + 2 by x − 7. Write your answer in the form P(x) = Q(x)· D(x) + R. To see the answer, pass your mouse over the colored area. x^{3} − 8x^{2} + x + 2 = (x^{2} − x − 6)(x − 7) − 40 The remainder theorem The value of a polynomial P(x) at x = a, P(a), is equal to the remainder upon dividing P(x) That is, when P(x) = Q(x)(x − a) + R, where Q(x) is the quotient and R is the remainder, then P(a) = R. For,
Example 2. Let f(x) = x^{3} − 3x^{2} − 13x + 15. We will use synthetic division to divide f(x) by x + 4. Now, what does the remainder theorem tell us? The value of f(x) at x = −4, is equal to the remainder: f(−4) = −45. Now let us divide f(x) by x − 5: What does the remainder theorem tell us here? f(5) = 0. But this means that 5 is a root of f(x) Moreover, since the remainder is 0 -- there is no remainder -- then (x − 5) is a factor of f(x). The synthetic division shows: x^{3} − 3x^{2} − 13x + 15 = (x^{2} + 2x − 3)(x − 5) This illustrates the Factor Theorem: The Factor Theorem. x − r is a factor of a polynomial P(x) if and only if r is a root of P(x). Problem 2. Let f(x) = x^{3} − 5x^{2} − 4x + 7. Use synthetic division to divide f(x) by x − 7. Therefore, according to the remainder theorem, f(7) = 77. Since the remainder is not 0 -- f(7) 0 -- upon dividing f(x) by x − 7, then (x − 7) is not a factor of f(x). And according to the factor theorem, 7 is not a root of f(x). Problem 3. Let g(x) = 3x ^{4} + 17x^{3} + 16x ^{2} − 10x + 4. Use synthetic division to divide g(x) by x + 2. According to the remainder theorem, g(−2) = 0. Therefore, what do you conclude about −2? −2 is a root of g(x). What do you conclude about (x + 2)? (x + 2) is a factor of g(x). Problem 4. Use synthetic division to divide x^{3} + 125 by x + 5. x^{3} + 125 = (x^{2} − 5x + 25)(x + 5) Next Topic: Roots of polynomials Copyright © 2018 Lawrence Spector Questions or comments? E-mail: themathpage@yandex.com Private tutoring available. |