Every topic on the board, the methods from the Lesson 5 slides, all worksheet questions (pp. 53–55) with full worked solutions, unlimited instant-check drills, and five 10-mark mock quizzes.
Write ax2 + bx + c as a product of two brackets (_x + _)(_x + _).
Method
Common factor first: take out the HCF of the numbers and the lowest power of x, e.g. 2x3 − 14x2 − 36x = 2x(x2 − 7x − 18). If a is negative, take out −1.
Identify a, b, c. Find two numbers m, n with mn = ac and m + n = b.
Split the middle term: ax2 + mx + nx + c.
Group in pairs and factor each pair; the brackets must match.
Take out the common bracket. Check by expanding.
Watch out
Product is a × c, not just c, when a ≠ 1.
Signs: if ac is negative, the two numbers have opposite signs.
"Factorise completely" means take the common factor out first and keep it in the answer.
Two-variable version works the same: x2 − 10xy + 9y2 = (x − y)(x − 9y).
Worked example · Slides 13–14
Factorise 6x2 − x − 2.
For 6x2 − x − 2: a = 6, b = −1, c = −2. Find two numbers with product ac = 6 × (−2) = −12 and sum b = −1 → 3 and −4.
Split the middle term: 6x2 − x − 2 = 6x2 + 3x − 4x − 2
A fast way to divide by a linear factor x − r, using only the coefficients.
Method
Write the divisor as x − r. So x + 3 gives r = −3, and x − 2 gives r = 2.
List the coefficients in descending powers, with 0 for missing terms.
Bring the first coefficient straight down.
Multiply it by r, write the result under the next coefficient, and add.
Repeat to the end. The last number is the remainder; the others are the quotient's coefficients, one degree lower.
Watch out
The sign of r is the opposite of the sign in the bracket.
In synthetic division you add; in long division you subtract.
A cubic divided by x − r gives a quadratic quotient.
Worked example · Slide 20
Use synthetic division to divide 2x3 + 4x2 − 5 by (x + 3). Write down the quotient and the remainder.
x + 3 = x − (−3), so r = −3.
Coefficients in descending powers (0 for each missing term): 2, 4, 0, −5
Bring down the first coefficient: 2. Then multiply by r and add, column by column: (−3) × 2 = −6, then 4 + (−6) = −2 (−3) × (−2) = 6, then 0 + 6 = 6 (−3) × 6 = −18, then −5 + (−18) = −23
−3
2
4
0
−5
−6
6
−18
2
−2
6
−23
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
70 questions: every worksheet question from pages 53–55, the slide examples, and some extras. Try each on paper first, then reveal the solution one step at a time.
1FactorisingSlide 13
Factorise 6x2 − x − 2.
Take out any common factor first. Then find two numbers that multiply to ac and add to b, split the middle term and factor by grouping.
For 6x2 − x − 2: a = 6, b = −1, c = −2. Find two numbers with product ac = 6 × (−2) = −12 and sum b = −1 → 3 and −4.
Split the middle term: 6x2 − x − 2 = 6x2 + 3x − 4x − 2
Use synthetic division to divide 2x3 + 4x2 − 5 by (x + 3). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x + 3 = x − (−3), so r = −3.
Coefficients in descending powers (0 for each missing term): 2, 4, 0, −5
Bring down the first coefficient: 2. Then multiply by r and add, column by column: (−3) × 2 = −6, then 4 + (−6) = −2 (−3) × (−2) = 6, then 0 + 6 = 6 (−3) × 6 = −18, then −5 + (−18) = −23
−3
2
4
0
−5
−6
6
−18
2
−2
6
−23
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 2x2 − 2x + 6, R = −23
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2SyntheticWS p.55 Q2a
Use synthetic division to divide 3x3 − 2x2 − 7x + 6 by (x + 1). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x + 1 = x − (−1), so r = −1.
Coefficients in descending powers: 3, −2, −7, 6
Bring down the first coefficient: 3. Then multiply by r and add, column by column: (−1) × 3 = −3, then −2 + (−3) = −5 (−1) × (−5) = 5, then −7 + 5 = −2 (−1) × (−2) = 2, then 6 + 2 = 8
−1
3
−2
−7
6
−3
5
2
3
−5
−2
8
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 3x2 − 5x − 2, R = 8
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3SyntheticWS p.55 Q2b
Use synthetic division to divide 2x3 − 5x − 7 by (x − 2). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x − 2, so r = 2.
Coefficients in descending powers (0 for each missing term): 2, 0, −5, −7
Bring down the first coefficient: 2. Then multiply by r and add, column by column: 2 × 2 = 4, then 0 + 4 = 4 2 × 4 = 8, then −5 + 8 = 3 2 × 3 = 6, then −7 + 6 = −1
2
2
0
−5
−7
4
8
6
2
4
3
−1
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 2x2 + 4x + 3, R = −1
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4SyntheticWS p.55 Q2c
Use synthetic division to divide 2x3 + 7x2 − 6x − 8 by (x + 4). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x + 4 = x − (−4), so r = −4.
Coefficients in descending powers: 2, 7, −6, −8
Bring down the first coefficient: 2. Then multiply by r and add, column by column: (−4) × 2 = −8, then 7 + (−8) = −1 (−4) × (−1) = 4, then −6 + 4 = −2 (−4) × (−2) = 8, then −8 + 8 = 0
−4
2
7
−6
−8
−8
4
8
2
−1
−2
0
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 2x2 − x − 2, R = 0
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5SyntheticWS p.55 Q2d
Use synthetic division to divide 3x3 − 5x2 + 4x + 2 by (x + 3). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x + 3 = x − (−3), so r = −3.
Coefficients in descending powers: 3, −5, 4, 2
Bring down the first coefficient: 3. Then multiply by r and add, column by column: (−3) × 3 = −9, then −5 + (−9) = −14 (−3) × (−14) = 42, then 4 + 42 = 46 (−3) × 46 = −138, then 2 + (−138) = −136
−3
3
−5
4
2
−9
42
−138
3
−14
46
−136
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 3x2 − 14x + 46, R = −136
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6SyntheticWS p.55 Q2e
Use synthetic division to divide x3 + 5 by (x + 2). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x + 2 = x − (−2), so r = −2.
Coefficients in descending powers (0 for each missing term): 1, 0, 0, 5
Bring down the first coefficient: 1. Then multiply by r and add, column by column: (−2) × 1 = −2, then 0 + (−2) = −2 (−2) × (−2) = 4, then 0 + 4 = 4 (−2) × 4 = −8, then 5 + (−8) = −3
−2
1
0
0
5
−2
4
−8
1
−2
4
−3
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = x2 − 2x + 4, R = −3
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7SyntheticWS p.55 Q2f
Use synthetic division to divide 5x3 − 3x2 − 6 by (x − 1). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x − 1, so r = 1.
Coefficients in descending powers (0 for each missing term): 5, −3, 0, −6
Bring down the first coefficient: 5. Then multiply by r and add, column by column: 1 × 5 = 5, then −3 + 5 = 2 1 × 2 = 2, then 0 + 2 = 2 1 × 2 = 2, then −6 + 2 = −4
1
5
−3
0
−6
5
2
2
5
2
2
−4
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 5x2 + 2x + 2, R = −4
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8SyntheticWS p.55 Q8.1
Use synthetic division to divide x2 + 3x − 7 by (x − 2). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x − 2, so r = 2.
Coefficients in descending powers: 1, 3, −7
Bring down the first coefficient: 1. Then multiply by r and add, column by column: 2 × 1 = 2, then 3 + 2 = 5 2 × 5 = 10, then −7 + 10 = 3
2
1
3
−7
2
10
1
5
3
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = x + 5, R = 3
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9SyntheticWS p.55 Q8.3
Use synthetic division to divide 4x2 + 10x − 9 by (x + 3). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x + 3 = x − (−3), so r = −3.
Coefficients in descending powers: 4, 10, −9
Bring down the first coefficient: 4. Then multiply by r and add, column by column: (−3) × 4 = −12, then 10 + (−12) = −2 (−3) × (−2) = 6, then −9 + 6 = −3
−3
4
10
−9
−12
6
4
−2
−3
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 4x − 2, R = −3
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10SyntheticExtraStretch
Use synthetic division to divide x4 − 3x3 + 5x − 2 by (x − 3). Write down the quotient and the remainder.
Write the divisor as x − r (so x + 3 gives r = −3). Bring down the first coefficient, then multiply by r and add, column by column.
x − 3, so r = 3.
Coefficients in descending powers (0 for each missing term): 1, −3, 0, 5, −2
Bring down the first coefficient: 1. Then multiply by r and add, column by column: 3 × 1 = 3, then −3 + 3 = 0 3 × 0 = 0, then 0 + 0 = 0 3 × 0 = 0, then 5 + 0 = 5 3 × 5 = 15, then −2 + 15 = 13
3
1
−3
0
5
−2
3
0
0
15
1
0
0
5
13
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = x3 + 5, R = 13
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1Remainder thmSlide 21
Use the Remainder Theorem to find the remainder when P(x) = 3x2 − x − 10 is divided by (x + 2).
Remainder Theorem: dividing by x − r leaves remainder P(r). Just substitute.
x + 2 = x − (−2), so r = −2. By the Remainder Theorem, the remainder is P(−2).
P(−2) = 3(−2)2 − (−2) − 10 = 12 + 2 − 10 = 4
Answer
R = P(−2) = 4
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2Remainder thmWS p.55 Q4
Use the Remainder Theorem to find the remainder when P(x) = 2x3 − 12x2 − x + 30 is divided by (x − 5).
Remainder Theorem: dividing by x − r leaves remainder P(r). Just substitute.
x − 5, so r = 5. By the Remainder Theorem, the remainder is P(5).
Use synthetic division to divide 2x3 − x2 − 4 by (x − 2). Write down the quotient and the remainder.
x − 2, so r = 2.
Coefficients in descending powers (0 for each missing term): 2, −1, 0, −4
Bring down the first coefficient: 2. Then multiply by r and add, column by column: 2 × 2 = 4, then −1 + 4 = 3 2 × 3 = 6, then 0 + 6 = 6 2 × 6 = 12, then −4 + 12 = 8
2
2
−1
0
−4
4
6
12
2
3
6
8
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 2x2 + 3x + 6, R = 8
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Question 4Remainder thm[1 mark]
Use the Remainder Theorem to find the remainder when P(x) = x3 − 3x2 + 5x − 1 is divided by (x − 1).
x − 1, so r = 1. By the Remainder Theorem, the remainder is P(1).
Use synthetic division to divide 3x3 + 4x2 − 5x + 1 by (x + 2). Write down the quotient and the remainder.
x + 2 = x − (−2), so r = −2.
Coefficients in descending powers: 3, 4, −5, 1
Bring down the first coefficient: 3. Then multiply by r and add, column by column: (−2) × 3 = −6, then 4 + (−6) = −2 (−2) × (−2) = 4, then −5 + 4 = −1 (−2) × (−1) = 2, then 1 + 2 = 3
−2
3
4
−5
1
−6
4
2
3
−2
−1
3
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 3x2 − 2x − 1, R = 3
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Question 4Remainder thm[1 mark]
Use the Remainder Theorem to find the remainder when P(x) = 2x3 − 5x2 + x + 4 is divided by (x − 3).
x − 3, so r = 3. By the Remainder Theorem, the remainder is P(3).
The remainder is 0, so x3 − 7x + 6 = (x − 2)(x2 + 2x − 3).
x2 + 2x − 3: product −3, sum 2 → 3 and −1, so x2 + 2x − 3 = (x + 3)(x − 1).
Answer
(x − 2)(x + 3)(x − 1)
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Question 3Synthetic[2 marks]
Use synthetic division to divide x4 − 2x3 + 3x − 4 by (x + 1). Write down the quotient and the remainder.
x + 1 = x − (−1), so r = −1.
Coefficients in descending powers (0 for each missing term): 1, −2, 0, 3, −4
Bring down the first coefficient: 1. Then multiply by r and add, column by column: (−1) × 1 = −1, then −2 + (−1) = −3 (−1) × (−3) = 3, then 0 + 3 = 3 (−1) × 3 = −3, then 3 + (−3) = 0 (−1) × 0 = 0, then −4 + 0 = −4
−1
1
−2
0
3
−4
−1
3
−3
0
1
−3
3
0
−4
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = x3 − 3x2 + 3x, R = −4
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Question 4Remainder thm[1 mark]
When P(x) = x3 + kx2 − 4x + 3 is divided by (x − 2) the remainder is 7. Find k.
Remainder Theorem: the remainder is P(2).
P(2) = 8 + 4k − 8 + 3 = 4k + 3
4k + 3 = 7 ⇒ k = 1
Answer
k = 1
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Question 5Factor thm[1 mark]
Use the Factor Theorem to show that (x + 1) is a factor of x5 + 1.
Factor Theorem: (x + 1) is a factor ⇔ P(−1) = 0, because x + 1 = x − (−1).
P(−1) = (−1)5 + 1 = −1 + 1 = 0
P(−1) = 0, so (x + 1)is a factor of P(x).
Answer
P(−1) = 0, so (x + 1)is a factor of P(x).
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Question 6Special products[1 mark]
Expand (x + 2)3 using a special product formula.
Use (A + B)3 = A3 + 3A2B + 3AB2 + B3 with A = x and B = 2.
(x + 2)3 = x3 + 3x2(2) + 3(x)(2)2 + 23
= x3 + 6x2 + 12x + 8
Answer
x3 + 6x2 + 12x + 8
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Question 1Factorising[2 marks]
Factorise completely −2x2 + 2x + 12.
Take out the common factor −2: −2x2 + 2x + 12 = −2(x2 − x − 6)
For x2 − x − 6: a = 1, b = −1, c = −6. Find two numbers with product ac = 1 × (−6) = −6 and sum b = −1 → 2 and −3.
Split the middle term: x2 − x − 6 = x2 + 2x − 3x − 6
So 4x3 − 8x2 + 5x − 72x − 3 = 2x2 − x + 1 − 42x − 3
Answer
Quotient Q(x) = 2x2 − x + 1, remainder R = −4
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Question 3Synthetic[2 marks]
Use synthetic division to divide 4x3 − 3x2 + 2x + 10 by (x + 1). Write down the quotient and the remainder.
x + 1 = x − (−1), so r = −1.
Coefficients in descending powers: 4, −3, 2, 10
Bring down the first coefficient: 4. Then multiply by r and add, column by column: (−1) × 4 = −4, then −3 + (−4) = −7 (−1) × (−7) = 7, then 2 + 7 = 9 (−1) × 9 = −9, then 10 + (−9) = 1
−1
4
−3
2
10
−4
7
−9
4
−7
9
1
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).
Answer
Q(x) = 4x2 − 7x + 9, R = 1
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Question 4Remainder thm[2 marks]
When P(x) = 2x3 + ax2 − 5x + 1 is divided by (x + 2), the remainder is 3. Find the value of a.
Use synthetic division to find the quotient when P(x) is divided by (x − 3).
[1]
x − 3, so r = 3.
Coefficients in descending powers: 2, −3, −11, 6
Bring down the first coefficient: 2. Then multiply by r and add, column by column: 3 × 2 = 6, then −3 + 6 = 3 3 × 3 = 9, then −11 + 9 = −2 3 × (−2) = −6, then 6 + (−6) = 0
3
2
−3
−11
6
6
9
−6
2
3
−2
0
Last number = remainder. The rest are the quotient's coefficients, one degree lower than P(x).