2.7 Polynomials, the factor theorem and algebraic division — Practise

Work through these on your own. To test a possible factor, substitute the root and check for zero; only then divide. For a division, always write the remainder, even when it is zero.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Divide $x^3 - 2x^2 + 4x - 7$ by $x - 1$ and give the quotient.
Find the remainder when $x^3 - 2x^2 + 4x - 7$ is divided by $x - 1$ .
Use the remainder theorem to find the remainder when $x^3 + 2x^2 - 5x + 1$ is divided by $x - 2$ .
Show that $x - 2$ is a factor of $p(x) = x^3 - 3x^2 + 4x - 4$ .
Factorise $x^3 - 6x^2 + 11x - 6$ completely.
Factorise $x^3 + 2x^2 - x - 2$ completely.
Divide $2x^3 + 3x^2 - 8x + 3$ by $x + 3$ . Give the quotient and state whether $x + 3$ is a factor.
Solve $x^3 - 4x^2 + x + 6 = 0$ .
Find the remainder when $x^4 - 3x + 2$ is divided by $x + 1$ .
Given that $x - 2$ is a factor of $x^3 + kx^2 - 4$ , find $k$ .
Factorise $x^3 - 2x^2 - 5x + 6$ completely.
Given that $x = 1$ is a root of $x^3 - 7x + 6 = 0$ , divide to find the other two roots.

Answers and concise solutions

1. The quotient

Answer: Divide: $x^3 - 2x^2 + 4x - 7 = (x - 1)(x^2 - x + 3) - 4$ . The quotient is $x^2 - x + 3$ .

2. The remainder

Answer: By the remainder theorem the remainder is $p(1) = 1 - 2 + 4 - 7 = -4$ .

3. The remainder

Answer: Substitute $x = 2$ : $2^3 + 2 \times 2^2 - 5 \times 2 + 1 = 8 + 8 - 10 + 1 = 7$ . The remainder is $7$ .

4. The value at the root

Answer: Substitute $x = 2$ : $p(2) = 8 - 12 + 8 - 4 = 0$ . Since $p(2) = 0$ , the factor theorem says $x - 2$ is a factor. In fact $p(x) = (x - 2)(x^2 - x + 2)$ .

5. Factorised

Answer: The constant is $-6$ , so try factors of $6$ . $p(1) = 1 - 6 + 11 - 6 = 0$ , so $x - 1$ is a factor. Dividing gives $(x - 1)(x^2 - 5x + 6)$ , and $x^2 - 5x + 6 = (x - 2)(x - 3)$ . So $x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$ .

6. Factorised

Answer: $p(1) = 1 + 2 - 1 - 2 = 0$ , so $x - 1$ is a factor. Dividing gives $(x - 1)(x^2 + 3x + 2)$ , and $x^2 + 3x + 2 = (x + 1)(x + 2)$ . So $x^3 + 2x^2 - x - 2 = (x - 1)(x + 1)(x + 2)$ .

7. The quotient

Answer: Divide: $2x^3 + 3x^2 - 8x + 3 = (x + 3)(2x^2 - 3x + 1)$ . The quotient is $2x^2 - 3x + 1$ . The remainder is zero, so $x + 3$ is a factor. Also $p(-3) = -54 + 27 + 24 + 3 = 0$ , which confirms it by the factor theorem.

8. The solutions

Answer: Try $x = -1$ : $-1 - 4 - 1 + 6 = 0$ , so $x + 1$ is a factor. Dividing gives $(x + 1)(x^2 - 5x + 6)$ , and $x^2 - 5x + 6 = (x - 2)(x - 3)$ . The roots are $x = -1$ , $x = 2$ and $x = 3$ .

9. The remainder

Answer: Write $x + 1$ as $x - (-1)$ , so substitute $x = -1$ : $(-1)^4 - 3 \times (-1) + 2 = 1 + 3 + 2 = 6$ . The remainder is $6$ .

10. The value of $k$

Answer: A factor gives a zero remainder, so $p(2) = 0$ : $2^3 + k \times 2^2 - 4 = 0$ , so $8 + 4k - 4 = 0$ and $4k = -4$ . Therefore $k = -1$ .

11. Factorised

Answer: $p(1) = 1 - 2 - 5 + 6 = 0$ , so $x - 1$ is a factor. Dividing gives $(x - 1)(x^2 - x - 6)$ , and $x^2 - x - 6 = (x + 2)(x - 3)$ . So $x^3 - 2x^2 - 5x + 6 = (x - 1)(x + 2)(x - 3)$ .

12. The quotient, then the roots

Answer: Since $x = 1$ is a root, $x - 1$ is a factor. Dividing gives $x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)$ . Factorise the quadratic: $x^2 + x - 6 = (x + 3)(x - 2)$ , so the other roots are $x = -3$ and $x = 2$ . The three roots are $-3$ , $1$ and $2$ .