Expanding binomials and taking out common factors – Year 9 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Multiply both terms by 5: \(5\times2x=10x\) and \(5\times(-3)=-15\). The answer is \(10x-15\). (1 mark)
   2. Multiply each term in the first bracket by each term in the second: \(x\times x=x^{2}\), \(x\times(-3)=-3x\), \(7\times x=7x\), \(7\times(-3)=-21\) (1 mark for the four products). Collect the middle terms, \(-3x+7x=4x\): \(x^{2}+4x-21\) (1 mark).
   3. The minus sign before the second bracket means multiply it by \(-1\): \(3x+6-x+4\). Collect: \(2x+10\). (1 mark)

   Check (b) with \(x=1\): \((1+7)(1-3)=8\times(-2)=-16\), and \(1+4-21=-16\). A grid method helps to avoid missing a product.

2. Answer to Question 2 (4 marks)

   1. The HCF of 12 and 18 is 6, so take 6 outside: \(12x\div6=2x\) and \(18\div6=3\). The answer is \(6(2x-3)\). (1 mark)
   2. Both terms contain x: \(x^{2}=x\times x\) and \(9x=9\times x\). Take x outside the bracket: \(x(x+9)\). (1 mark)
   3. The HCF of 15 and 10 is 5, and both terms contain a, so the HCF is \(5a\) (1 mark). Dividing: \(15ab\div5a=3b\) and \(10a^{2}\div5a=2a\). The answer is \(5a(3b+2a)\), also written \(5a(2a+3b)\) (1 mark).

   A partial factorisation such as \(5(3ab+2a^{2})\) earns only 1 mark in (c). Always check that nothing more can come out of the bracket; then expand to check.

3. Answer to Question 3 (4 marks)

   1. Write the square as two brackets: \((x+5)(x+5)=x^{2}+5x+5x+25\) (1 mark). The two middle terms add to \(10x\): \(x^{2}+10x+25\) (1 mark). No marks are given for \(x^{2}+25\), which forgets the middle terms.
   2. Expand: \(4x^{2}-6x+6x-9\). The middle terms cancel, leaving \(4x^{2}-9\). This is a difference of two squares, \((2x)^{2}-3^{2}\). (1 mark)
   3. Since \(64=8^{2}\), use \(a^{2}-b^{2}=(a+b)(a-b)\): \((x+8)(x-8)\). (1 mark)

   Check (a) with \(x=1\): \(6^{2}=36\) and \(1+10+25=36\). Check (c) by expanding: \(x^{2}-8x+8x-64=x^{2}-64\).

4. Answer to Question 4 (4 marks)

   1. Area = length × width \(=(x+5)(x+3)\). Expanding gives \(x^{2}+3x+5x+15\) (1 mark). The middle terms add to \(8x\), so the area is \(x^{2}+8x+15\), as required (1 mark).
   2. Substitute \(x=6\): \(6^{2}+8\times6+15=36+48+15=\)99 m². A quicker check uses the sides: \(6+5=11\) m and \(6+3=9\) m, and \(11\times9=99\). (1 mark)
   3. Multiply the area by the price per square metre: \(99\times2.50=\)£247.50. (1 mark)

   In a "show that" question every step must be written, because the answer is already given.

5. Answer to Question 5 (4 marks)

   1. Follow the steps in order. Adding 3 gives \(n+3\); multiplying by 4 gives \(4(n+3)\) (1 mark); subtracting 12 gives \(4(n+3)-12\); dividing by the starting number gives \(\dfrac{4(n+3)-12}{n}\) (1 mark).
   2. Expand the numerator: \(4n+12-12=4n\), because the 12s cancel (1 mark). Then \(\dfrac{4n}{n}=4\), since n divides exactly when \(n\ne0\) (1 mark). The result is 4 for every allowed starting number, so Sam is right.

   Testing with \(n=7\): \(7+3=10\), \(10\times4=40\), \(40-12=28\), \(28\div7=4\). A numerical example shows the trick working once, but only the algebra proves it always works. The condition \(n\ne0\) matters because dividing by zero is undefined.

Total : 20 marks