S2 Mathematics Practice Book · Unit 2: Polynomials

Unit 2 Polynomials

Key unit competence: Solve problems involving operations and factorisation of polynomials.

Key ideas

A polynomial in $x$ is a sum of terms $a_{n} x^{n} + \dots + a_{1} x + a_{0}$ with whole-number powers. The highest power is the degree; $a_{n}$ is the leading coefficient; $a_{0}$ the constant term.
By degree: constant (0), linear (1), quadratic (2), cubic (3). By number of terms: monomial, binomial, trinomial.
Add/subtract by collecting like terms; multiply each term of one polynomial by each term of the other.
Euclidean (long) division: $P (x) = D (x) Q (x) + R (x)$ with degree of $R$ less than degree of $D$ . Synthetic division is a quick method for dividing by $x - a$ .
Remainder theorem: the remainder when $P (x)$ is divided by $x - a$ is $P (a)$ . Factor theorem: $x - a$ is a factor if and only if $P (a) = 0$ .
Two polynomials are equal when their corresponding coefficients are equal. Composite: $(f \circ g) (x) = f (g (x))$ . The inverse of $f (x) = a x + b$ is $f^{- 1} (x) = \frac{x - b}{a}$ .
Identities: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ , $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ , $(a + b) (a - b) = a^{2} - b^{2}$ .
Factorisation methods: common factor, grouping, identities, sum and product (for $x^{2} + b x + c$ find numbers with sum $b$ and product $c$ ), and zeros of the polynomial.

Formulas to know

(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}

a^{2} - b^{2} = (a - b) (a + b)

P (x) = D (x) Q (x) + R (x)

P (a) = 0 ⟺ (x - a) is a factor

Common misconceptions

" $(a + b)^{2} = a^{2} + b^{2}$ ." No: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .
"The remainder of $P (x) \div (x + 2)$ is $P (2)$ ." No: $x + 2 = x - (- 2)$ , so the remainder is $P (- 2)$ .
" $(f \circ g) (x) = f (x) \times g (x)$ ." No: it means substitute $g (x)$ into $f$ .

Worked examples

Example 1. Find the remainder when $P (x) = x^{3} - 2 x^{2} + 4 x - 5$ is divided by $x - 2$ .

Solution

$P (2) = 8 - 8 + 8 - 5 = 3$ . The remainder is 3.

Example 2. Factorise $x^{2} + 5 x + 6$ .

Solution

Two numbers with sum 5 and product 6: 2 and 3. So $x^{2} + 5 x + 6 = (x + 2) (x + 3)$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

State the degree, leading coefficient and constant term of $P (x) = - 2 x^{4} + 3 x^{2} - x + 7$ .
Show answer
Degree 4, leading coefficient $- 2$ , constant term 7.
Common mistake: Taking the first number written as the constant term.
Factorise $x^{2} - x - 20$ using sum and product.
Show answer
Sum $- 1$ , product $- 20$ : $- 5$ and 4. $(x - 5) (x + 4)$ .
Common mistake: Choosing $5$ and $- 4$ (sum $+ 1$ ).
Expand and simplify $(x + 2) (x^{2} - 3 x + 1)$ .
Show answer
$x^{3} - 3 x^{2} + x + 2 x^{2} - 6 x + 2 = x^{3} - x^{2} - 5 x + 2$ .
Common mistake: Multiplying only the first terms.
Find $a$ and $b$ if $x^{2} + a x + b$ has zeros 2 and $- 5$ .
Show answer
$(x - 2) (x + 5) = x^{2} + 3 x - 10$ , so $a = 3$ , $b = - 10$ .
Common mistake: Writing the factors as $(x + 2) (x - 5)$ .
Use long division to divide $x^{3} + 2 x^{2} - 5 x - 6$ by $x + 1$ .
Show answer
Quotient $x^{2} + x - 6$ , remainder 0. So $x^{3} + 2 x^{2} - 5 x - 6 = (x + 1) (x - 2) (x + 3)$ .
Common mistake: Sign errors when subtracting each line.
Given $f (x) = x^{2} + 1$ and $g (x) = x - 3$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$ .
Show answer
$f (g (x)) = (x - 3)^{2} + 1 = x^{2} - 6 x + 10$ ; $g (f (x)) = x^{2} - 2$ . They are different.
Common mistake: Assuming composition is commutative.
Show that $x + 3$ is a factor of $x^{3} + 3 x^{2} - 4 x - 12$ and factorise completely.
Show answer
$P (- 3) = - 27 + 27 + 12 - 12 = 0$ . Grouping: $x^{2} (x + 3) - 4 (x + 3) = (x + 3) (x - 2) (x + 2)$ .
Common mistake: Evaluating $P (3)$ instead of $P (- 3)$ .
Factorise $3 x^{2} - 12$ .
Show answer
$3 (x^{2} - 4) = 3 (x - 2) (x + 2)$ .
Common mistake: Stopping at $3 (x^{2} - 4)$ .

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Factorise $x^{2} + 7 x + 12$ .	a. $k = 7$
2. $a x^{2} + b x + 3 = 2 x^{2} - 5 x + c$ for all $x$ . Find $a$ , $b$ , $c$ .	b. $a = 2, b = - 5, c = 3$
3. $f (x) = x^{2} - 1$ and $g (x) = 2 x + 3$ . Find $(g \circ f) (2)$ .	c. $(x + 3) (x + 4)$
4. $x - 2$ is a factor of $x^{3} - k x + 6$ . Find $k$ .	d. 9

Show answer

1 → c ( $(x + 3) (x + 4)$ ) · 2 → b ( $a = 2, b = - 5, c = 3$ ) · 3 → d (9) · 4 → a ( $k = 7$ )

$3 + 4 = 7$ and $3 \times 4 = 12$ .
Equate corresponding coefficients.
Step 1. First $f (2) = 4 - 1 = 3$ .
Step 2. Then $g (3) = 2 (3) + 3 = 9$ .
Step 1. Factor theorem: $P (2) = 0$ , so $8 - 2 k + 6 = 0$ .
Step 2. $14 = 2 k$ , so $k = 7$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$x = 2$ · $(x + 1) (x + 3)$ · $(x - 1) (x - 3) (x + 2)$ · $(2 x + 1) (x + 3)$
1. Factorise $2 x^{2} + 7 x + 3$ . →
2. Factorise $x^{3} - 2 x^{2} - 5 x + 6$ completely. →
3. If $f (x) = x + 4$ and $g (x) = 2 x - 1$ , solve $(g \circ f) (x) = 11$ . →
Show answer
1: $(2 x + 1) (x + 3)$ · 2: $(x - 1) (x - 3) (x + 2)$ · 3: $x = 2$
- Product $2 \times 3 = 6$ and sum 7 give 6 and 1: $2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + (x + 3)$ .
- Step 1. $P (1) = 1 - 2 - 5 + 6 = 0$ , so $x - 1$ is a factor: $x^{3} - 2 x^{2} - 5 x + 6 = (x - 1) (x^{2} - x - 6)$ .
  Step 2. $x^{2} - x - 6 = (x - 3) (x + 2)$ .
- Step 1. $g (f (x)) = 2 (x + 4) - 1 = 2 x + 7 = 11$ .
  Step 2. So $x = 2$ .
Matching
Match each item in column A with the correct answer in column B. Write the letter next to each number.
Column A Column B
1. Factorise $2 x^{2} + 5 x - 3$ . a. $3 x (2 x + 3)$
2. If $f (x) = 2 x + 1$ and $g (x) = x^{2}$ , find $(f \circ g) (x)$ . b. $2 x^{2} + 1$
3. Factorise $6 x^{2} + 9 x$ . c. $x^{2} + x - 6$
4. Expand $(x + 3) (x - 2)$ . d. $(2 x - 1) (x + 3)$
Show answer
1 → d ( $(2 x - 1) (x + 3)$ ) · 2 → b ( $2 x^{2} + 1$ ) · 3 → a ( $3 x (2 x + 3)$ ) · 4 → c ( $x^{2} + x - 6$ )
- Step 1. $a c = - 6$ ; find two numbers with product $- 6$ and sum 5: 6 and $- 1$ .
  Step 2. $2 x^{2} + 6 x - x - 3 = 2 x (x + 3) - (x + 3) = (2 x - 1) (x + 3)$ .
- $f (g (x)) = 2 (x^{2}) + 1$ .
- The highest common factor is $3 x$ .
- $x^{2} - 2 x + 3 x - 6 = x^{2} + x - 6$ .
Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$x^{2} + x - 6$ · $f^{- 1} (x) = \frac{x + 6}{3}$ · $(x - 9) (x + 1)$ · $(x - 3)^{2}$
1. Divide $x^{3} - 7 x + 6$ by $x - 1$ (synthetic division). What is the quotient? →
2. Factorise $x^{2} - 6 x + 9$ . →
3. Find the inverse of $f (x) = 3 x - 6$ . →
Show answer
1: $x^{2} + x - 6$ · 2: $(x - 3)^{2}$ · 3: $f^{- 1} (x) = \frac{x + 6}{3}$
- Coefficients $1, 0, - 7, 6$ with $a = 1$ give $1, 1, - 6$ and remainder 0.
- Perfect square: $x^{2} - 2 (3) x + 3^{2}$ .
- $y = 3 x - 6 \Rightarrow x = \frac{y + 6}{3}$ .

Column A	Column B
1. Factorise $2 x^{2} + 5 x - 3$ .	a. $3 x (2 x + 3)$
2. If $f (x) = 2 x + 1$ and $g (x) = x^{2}$ , find $(f \circ g) (x)$ .	b. $2 x^{2} + 1$
3. Factorise $6 x^{2} + 9 x$ .	c. $x^{2} + x - 6$
4. Expand $(x + 3) (x - 2)$ .	d. $(2 x - 1) (x + 3)$

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