Worked examples · Sheet 3

Ruffini's rule and the remainder theorem

Theory used: Remainder theorem Ruffini's theorem Ruffini's rule

Exercise 1

Compute, without carrying out the divisions, the remainder of the division of $P (x) = x^{4} - 3 x^{3} + 2 x - 1$ by $x - 2$ and by $x + 1$ .

P (2) = 2^{4} - 3 \cdot 2^{3} + 2 \cdot 2 - 1 = 16 - 24 + 4 - 1 = - 5

P (- 1) = (- 1)^{4} - 3 \cdot (- 1)^{3} + 2 \cdot (- 1) - 1 = 1 + 3 - 2 - 1 = 1

The remainders are $- 5$ and $1$ .

By the remainder theorem, the remainder of the division of $P (x)$ by $x - c$ is the value $P (c)$ , so a single evaluation of the polynomial is enough. In the first division the divisor is $x - 2$ , therefore $c = 2$ and the remainder is $P (2) = - 5$ .

The divisor $x + 1$ is written as $x - (- 1)$ , so $c = - 1$ and the remainder is $P (- 1) = 1$ . Since both remainders are nonzero, by Ruffini’s theorem neither $x - 2$ nor $x + 1$ divides $P$ .

Exercise 2

Using Ruffini’s rule, find the quotient and the remainder of the division of $P (x) = 2 x^{3} - 3 x^{2} + 4$ by $x - 2$ .

2 x^{3} - 3 x^{2} + 0 x + 4

222 - 3 41 022448

Q (x) = 2 x^{2} + x + 2, R = 8

The polynomial lacks the term of degree $1$ , so in the first row of the scheme the corresponding coefficient is $0$ ; the number on the left is $c = 2$ , because the divisor is $x - 2$ .

The first coefficient $2$ comes down unchanged. Then the multiply-and-add step is repeated: $2 \cdot 2 = 4$ and $- 3 + 4 = 1$ ; then $1 \cdot 2 = 2$ and $0 + 2 = 2$ ; finally $2 \cdot 2 = 4$ and $4 + 4 = 8$ . The first three numbers in the row of results are the coefficients of the quotient, which has degree $2$ , and the last one is the remainder.

The remainder theorem provides the check: $P (2) = 16 - 12 + 4 = 8$ , which equals the remainder in the scheme.

Exercise 3

Find $k$ so that $x + 2$ divides $P (x) = x^{3} + k x^{2} + x - 6$ , and write the division for the value found.

P (- 2) = - 8 + 4 k - 2 - 6 = 4 k - 16

4 k - 16 = 0 ⟹ k = 4

For $k = 4$ , with Ruffini’s rule:

- 2 11 4 - 2 2 1 - 4 - 3 - 6 60

x^{3} + 4 x^{2} + x - 6 = (x + 2) (x^{2} + 2 x - 3)

By Ruffini’s theorem, $x + 2$ divides $P$ if and only if $P (- 2) = 0$ . The value $P (- 2)$ depends on $k$ and equals $4 k - 16$ , so the divisibility condition becomes the equation $4 k - 16 = 0$ , whose solution is $k = 4$ .

With $k = 4$ Ruffini’s scheme with $c = - 2$ proceeds as follows: the coefficient $1$ comes down; $1 \cdot (- 2) = - 2$ and $4 - 2 = 2$ ; $2 \cdot (- 2) = - 4$ and $1 - 4 = - 3$ ; $- 3 \cdot (- 2) = 6$ and $- 6 + 6 = 0$ . The remainder is $0$ , as required, and the quotient is $x^{2} + 2 x - 3$ , therefore the division provides the factorization shown.