WJEC MATHEMATICS INTERMEDIATE ALGEBRA EXPANDING 1 Contents Single Brackets (Santa's hat) Double Brackets (Foil) Multiple brackets with simplifying Credits WJEC Question bank http://www.wjec.co.uk/question-bank/question-search.html 2 Expanding brackets means that you want to remove brackets from an expression. There are two main types; Remember: π₯ + π₯ = 2π₯ and π₯ × π₯ = π₯2 Single Brackets A single bracket question looks like this: 3(2π₯ + 4) Which means 3 multiplied by 2π₯ and multiplied by 4. So, to expand this, we need to multiply the 3 and the 2π₯ and then multiply the 3 and the 4. We can add multiply lines to show this: 3(2π₯ + 4) Look familiar? This method is sometimes called the Santa's hat method So, the answer to the above example: 6π₯ + 12 3 × 2π₯ 3×4 3 Exercise A7 1. 5(6π¦ + 2) 2. 7(3π + 8) 3. 4(3π₯ + 3) 4. 12(9π + 8) 5. π₯(6π₯ + 8) Negative Numbers Most marks are lost on these questions when people become confused with negative signs. Consider this example: 2(5π₯ β 7) = 10π₯ β 14 2 × 5π₯ 2 × β7 Consider this more difficult example: β4(9π₯ β 7) Remember: Multiplying two negative numbers results in = β36π₯ + 28 β4 × 9π₯ a positive number β4 × β7 4 Exercise A8 Expand: 1. 4(3π¦ β 2) 2. 9(5π β 2) 3. β5(8π₯ β 11) 4. β12(8π β 11) 5. βπ₯(4π₯ β 0.5) Double Brackets Double brackets questions typically look like this: (π₯ + 3)(π₯ + 5) To expand this, we need to multiply every term in the first bracket by every term in the second bracket. We can draw multiplication lines on as follows: (π₯ + 3)(π₯ + 5) We need to multiply the first term in each bracket, the outside term of each bracket, the inside term in each bracket and the last term in each bracket. FIRST - OUTSIDE - INSIDE - LAST (Or FOIL for short) 5 (π₯ + 3)(π₯ + 5) FIRST (π₯ + 3)(π₯ + 5) OUTSIDE (π₯ + 3)(π₯ + 5) INSIDE (π₯ + 3)(π₯ + 5) LAST Going back to the earlier example: (π₯ + 3)(π₯ + 5) Expanding, we get: π₯ 2 + 5π₯ + 3π₯ + 15 FIRST π₯×π₯ OUTSIDE π₯×5 INSIDE 3×π₯ LAST 3×5 Notice how the middle two terms can be simplified. If you are unsure of this, see the Simplifying booklet. So, we get π₯ 2 + 8π₯ + 15 Again, take care with negative signs (π₯ β 4)(π₯ β 7) = π₯ 2 β 7π₯ β 4π₯ + 28 FIRST π₯×π₯ OUTSIDE π₯ × β7 INSIDE β4 × π₯ LAST β4 × β7 6 = π₯ 2 β 11π₯ + 28 Exercise A9 Expand: 1. (π₯ + 4)(π₯ + 9) 2. (π₯ + 3)(π₯ + 8) 3. (π₯ + 1)(π₯ + 4) 4. (π₯ + 3)(π₯ + 7) 5. (π₯ + 2)(π₯ + 8) 6. (π₯ + 9)(π₯ + 4) 7. (π₯ + 2)(π₯ + 5) 8. (π₯ + 6)(π₯ + 2) 9. (π₯ + 3)(π₯ + 8) 10. (π₯ β 4)2 Remember: 'Squared' means multiplied by itself 7 Multiple Brackets and Simplifying Lots of questions at GCSE require you to expand brackets and then simplify. If you get stuck on the simplifying step, see the booklet 'Simplifying'. Example Consider this question. It's clearly a question that required two Santa's hats! 2(4π₯ β 3) β 7(2π₯ + 5) = 8π₯ β 6 β 14π₯ β 35 2×4π₯ 2×β3 β7×2π₯ β7×+5 = β6π₯ β 41 Example 2 (Secret Santa's hat) On first glance, this question does not seem like a Santa's hat question but there is a sign in front of the second bracket. This minus sign acts on the 2π₯ and on the β8π¦. We can show this with multiplication lines. (4π₯ + 2π¦) β (2π₯ β 8π¦) = 4π₯ + 2π¦ β 2π₯ + 8π¦ This simplifies to; β(β8) = 2π₯ + 10π¦ 8 Exercise A10 Expand and simplify: 1. 2(9π₯ β 3) β 5(6π₯ + 9) 2. 7(12π¦ β 4) β 4(2π¦ + 11) 3. 4(8π β 6) β 7(9π + 7) 4. 11(6π₯ β 9π¦) β (π₯ + 2π¦) 5. 9(6π₯ β π) β (7π₯ + 9π) Exam Questions A3 1. 2. 3. 4. 9 5. 6. 7. 8. 9. 10
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