Fill in the blanks to complete the following identities:
- (i) s2 − 11s + 24 = (________) (________)
- (ii) (________) (x + 1) = (3x2 − 4x − 7)
- (iii) 10x2 − 11x − 6 = (2x − ___) (___ + 2)
- (iv) 6x2 + 7x + 2 = (____________) (____________)
Step-by-step solution
Idea: Use (x + a)(x + b) = x2 + (a + b)x + ab: find two numbers whose product is the constant term and whose sum is the coefficient of the middle term. When the x2 term has a number in front (like 6x2), split the middle term into two parts whose product equals (first coefficient × constant), then group. Always multiply your brackets back out to check.
(i) s2 − 11s + 24 = (________) (________)
- Compare with s2 + (a + b)s + ab: we need a + b = −11 and ab = 24.
- The product is positive and the sum is negative, so both numbers are negative. Pairs for 24: (−1, −24), (−2, −12), (−3, −8), (−4, −6). Only −3 + (−8) = −11 ✓½ mark
- s2 − 11s + 24 = (s − 3)(s − 8)½ mark
(ii) (________) (x + 1) = (3x2 − 4x − 7)
- The missing bracket must start with 3x (because 3x × x = 3x2) and end with a number k such that k × 1 = −7. So k = −7 and the bracket is (3x − 7).½ mark
- Check the middle term: (3x − 7)(x + 1) = 3x2 + 3x − 7x − 7 = 3x2 − 4x − 7 ✓½ mark
(iii) 10x2 − 11x − 6 = (2x − ___) (___ + 2)
- First terms: 2x × (?) = 10x2, so the second blank is 5x.
- Last terms: (−?) × 2 = −6, so the first blank is 3.½ mark
- Check the middle term: (2x − 3)(5x + 2) = 10x2 + 4x − 15x − 6 = 10x2 − 11x − 6 ✓½ mark
(iv) 6x2 + 7x + 2 = (____________) (____________)
- Multiply the first coefficient by the constant: 6 × 2 = 12. Find two numbers with product 12 and sum 7: 3 and 4.
- Split the middle term: 6x2 + 7x + 2 = 6x2 + 3x + 4x + 2½ mark
- Group: 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1)½ mark
- Check: (3x + 2)(2x + 1) = 6x2 + 3x + 4x + 2 = 6x2 + 7x + 2 ✓
Check: Put x = 1 (or s = 1). (i) 1 − 11 + 24 = 14 and (−2)(−7) = 14 ✓. (ii) 3 − 4 − 7 = −8 and (−4)(2) = −8 ✓. (iii) 10 − 11 − 6 = −7 and (−1)(7) = −7 ✓. (iv) 6 + 7 + 2 = 15 and (5)(3) = 15 ✓.
Answer to write in the exam
(i)
s2 − 11s + 24 = s2 + (−3 − 8)s + (−3)(−8) [(x + a)(x + b) = x2 + (a + b)x + ab]
∴ s2 − 11s + 24 = (s − 3)(s − 8)
(ii)
(3x − 7)(x + 1) = 3x2 + 3x − 7x − 7 = 3x2 − 4x − 7
∴ (3x − 7)(x + 1) = 3x2 − 4x − 7
(iii)
(2x − 3)(5x + 2) = 10x2 + 4x − 15x − 6 = 10x2 − 11x − 6
∴ 10x2 − 11x − 6 = (2x − 3)(5x + 2)
(iv)
6x2 + 7x + 2 = 6x2 + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
∴ 6x2 + 7x + 2 = (3x + 2)(2x + 1)
Common mistakes that cost marks
- In (i), choosing +3 and +8: their product is 24 but their sum is +11, giving s2 + 11s + 24. The middle term is −11s, so both numbers must be negative.
- In (iii), filling (2x − 2)(3x + 2) or similar without checking. The first terms must multiply to 10x2 and the last terms to −6; then the middle term must come out as −11x.
- Not multiplying back. Writing the brackets in a different order, such as (2x + 1)(3x + 2) in (iv), is fine, but always expand once to confirm.
How this can come in the exam
x2 − 2x − 15 is equal to
- (x + 5)(x − 3)
- (x − 5)(x + 3)
- (x − 15)(x + 1)
- (x − 5)(x − 3)
Show answer
(B) (x − 5)(x + 3)
We need product −15 and sum −2: −5 and +3. So (x − 5)(x + 3).
Fill in the blanks: 2x2 + 7x + 3 = (2x + ___)(x + ___).
Show answer
2 × 3 = 6; two numbers with product 6 and sum 7 are 1 and 6 (½ mark). 2x2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) (1 mark) = (2x + 1)(x + 3) (½ mark).Try one yourself
Fill in the blanks: y2 + 5y − 14 = (________)(________).
Show answer
Product −14, sum 5: 7 and −2. (y + 7)(y − 2).
More questions like this
- Select and use the identity that will help you to find the following products without multiplying directly:
- Factor the following:
- James and Reshma were talking about algebraic identities they learnt in school.
James: (a − b)2 (a + b) = (a2 − 2ab + b2)(a + b)
Reshma: I have a different idea. (a − b)2 (a + b) = (a − b) [(a − b) (a + b)] = (a − b)(a2 − b2)
I will find this product to get the answer.
According to you, who is correct and why?
Try to combine more such identities and find new results. - What do you think (a + b)3 will look like?
- What if we have a cube of edge a + b? Can we divide a cube of edge (a + b) into smaller cubes and cuboids and represent this new identity?