Mathematics • Year 10 • Unit 2 • Lesson 3
Expanding Binomials — Mixed Challenge
Pull together every idea from Lesson 3: FOIL, perfect squares (a ± b)², difference patterns and coefficients on x. Pick the right tool for each problem, find another student's slip, and design two binomials whose product hits a target.
1. Mixed problems — choose the right tool
Show your working. 3 marks each
1.1 Expand (x + 7)(x + 2).
1.2 Expand (x − 4)(x + 6).
1.3 Expand (3x + 2)(x − 5).
1.4 Expand (x + 8)(x − 8). What is unusual about the middle term?
1.5 Expand (3x − 2)². Use the perfect-square pattern.
1.6 A rectangle has area (x + 4)(x + 7). (a) Expand. (b) Find the area when x = 3. (c) What are the side lengths when x = 3?
2. Find the mistake
Another Year 10 student has tried to expand (x − 5)². Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — expand (x − 5)²:
Line 1: (x − 5)² = (x − 5)(x − 5)
Line 2: F: x · x = x²; O: x · (−5) = −5x; I: (−5) · x = −5x; L: (−5)(−5) = −25
Line 3: Sum: x² + (−5x) + (−5x) + (−25)
Line 4: = x² − 10x − 25
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? What is (−5) × (−5)? Two negatives don't give a negative.3. Open-ended challenge — design a binomial product
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design two binomials of the form (x + a)(x + b) — where a and b are whole numbers (possibly negative) — whose product, when expanded, has middle coefficient = 7 and constant = 10. (i.e. the expansion looks like x² + 7x + 10.)
In your submission, include:
(i) Your two binomials.
(ii) Your full FOIL working showing the expansion equals x² + 7x + 10.
(iii) A one-sentence note: which two whole numbers a and b multiply to 10 AND add to 7?
Bonus: Design a second pair whose product is x² − 7x + 10. (Now the two factors should multiply to +10 but add to −7.)
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (x + 7)(x + 2)
F: x²; O: 2x; I: 7x; L: 14. Sum: x² + 9x + 14.
1.2 — (x − 4)(x + 6)
F: x²; O: 6x; I: −4x; L: −24. Sum: x² + 6x − 4x − 24 = x² + 2x − 24.
1.3 — (3x + 2)(x − 5)
F: 3x²; O: −15x; I: 2x; L: −10. Sum: 3x² − 15x + 2x − 10 = 3x² − 13x − 10.
1.4 — (x + 8)(x − 8)
F: x²; O: −8x; I: 8x; L: −64. Sum: x² − 8x + 8x − 64 = x² − 64.
The middle term cancels (−8x + 8x = 0) — a "difference of squares" pattern, the Lesson 5 topic.
1.5 — (3x − 2)²
= (3x − 2)(3x − 2). F: 9x²; O: −6x; I: −6x; L: +4. Sum: 9x² − 12x + 4.
1.6 — Rectangle area (x + 4)(x + 7)
(a) FOIL: x² + 7x + 4x + 28 = x² + 11x + 28.
(b) When x = 3: 9 + 33 + 28 = 70 square units.
(c) Side lengths: x + 4 = 7 and x + 7 = 10. Check: 7 × 10 = 70 ✓.
2 — Find the mistake
(a) The mistake is on Line 2 (and it propagates to Lines 3 and 4).
(b) The Last product (−5)(−5) should equal +25, not −25 — a negative times a negative is always positive. The student dropped the sign flip.
(c) Corrected working:
(x − 5)² = (x − 5)(x − 5)
F: x²; O: −5x; I: −5x; L: +25
Sum: x² + (−5x) + (−5x) + 25
= x² − 10x + 25.
The perfect-square pattern (a − b)² = a² − 2ab + b² always produces a positive last term.
3 — Open-ended challenge (sample solutions)
Main: produce x² + 7x + 10. We need a × b = 10 and a + b = 7. Pair: 2 and 5.
Binomials: (x + 2)(x + 5).
FOIL: x² + 5x + 2x + 10 = x² + 7x + 10 ✓.
Bonus: produce x² − 7x + 10. Need product +10 and sum −7. Pair: −2 and −5.
Binomials: (x − 2)(x − 5).
FOIL: x² − 5x − 2x + 10 = x² − 7x + 10 ✓.
The pattern is general: in (x + a)(x + b) the constant is ab and the middle coefficient is a + b — that observation is the engine of Lesson 6 factorising trinomials.
Marking: 1 for identifying the (2,5) pair, 2 for correct FOIL working that reaches x² + 7x + 10, 1 for the explanation. Bonus pair earns recognition only — not extra marks.