Algebraic notation, expanding and factorising – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(4\times3-(-2)=12+2=\) 14 We replace \(a\) by 3 and \(b\) by \((-2)\). Subtracting a negative number is the same as adding, so \(4a-b=14\).
   2. \(3\times(-2)+5=-6+5=\) −1 Multiply first: a positive times a negative gives a negative, \(-6\). Then add 5, so \(ab+c=-1\).
   3. \(3^2-(-2)^2=9-4=\) 5 Both squares are positive because \((-2)^2=4\), not \(-4\). So \(a^2-b^2=5\).
   4. \(2\times25-3\times3\times(-2)=50+18=\) 68 Square before multiplying: \(c^2=25\). The product \(3ab=-18\), and subtracting \(-18\) adds 18, so \(2c^2-3ab=68\).

2. Answer to Question 2

   1. \(5x-2x=3x\) and \(3y+4y=7y\), so \(3x+7y\) Only terms with exactly the same letters can be collected.
   2. \(3a^2-a^2=2a^2\) and \(2a+7a=9a\), so \(2a^2+9a\) \(a^2\) and \(a\) are unlike terms, so they stay separate.
   3. \(4pq-pq=3pq\) and \(3p+p=4p\), so \(3pq+4p\) \(pq\) and \(p\) are different terms, so the answer has two terms.
   4. The terms \(2x\), \(3y\) and \(5\) are all unlike, so part (d) cannot be simplified. Only like terms can be added together.

3. Answer to Question 3

   1. \(5\times2x=10x\) and \(5\times(-3)=-15\), so \(10x-15\) Each term inside the bracket is multiplied by 5.
   2. \(-3\times x=-3x\) and \(-3\times4=-12\), so \(-3x-12\) The negative sign belongs to the 3, so both terms become negative.
   3. \(x^3-7x\) Multiply each term by \(x\): \(x\times x^2=x^3\) and \(x\times(-7)=-7x\), so the answer is \(x^3-7x\).
   4. \(4x+8-3x+3=\) \(x+11\) Note that \(-3\times(-1)=+3\). Then collect: \(4x-3x=x\) and \(8+3=11\), giving \(x+11\).

4. Answer to Question 4

   1. \(x^2+5x+4x+20=\) \(x^2+9x+20\) Every term in the first bracket multiplies every term in the second, then the like terms \(5x\) and \(4x\) are collected.
   2. \(x^2+8x-3x-24=\) \(x^2+5x-24\) The middle terms give \(8x-3x=5x\), and \(-3\times8=-24\).
   3. \(6x^2-8x+3x-4=\) \(6x^2-5x-4\) Here \(2x\times3x=6x^2\) and \(1\times(-4)=-4\); the middle terms give \(-8x+3x=-5x\).
   4. \((x-6)(x-6)=x^2-6x-6x+36=\) \(x^2-12x+36\) A squared bracket means the bracket times itself, so there are two middle terms, not one: a common mistake is to write \(x^2+36\).

5. Answer to Question 5

   1. HCF is 6: \(6(2x+3)\) Check: \(6\times2x+6\times3=12x+18\). So \(12x+18=6(2x+3)\).
   2. HCF is \(x\): \(x(x-9)\) Both terms contain \(x\), so \(x\) is taken outside. Check: \(x\times x-9x=x^2-9x\).
   3. HCF is \(3ab\): \(3ab(2a+5b)\) The HCF of 6 and 15 is 3, and both terms contain \(a\) and \(b\). Check: \(3ab\times2a=6a^2b\) and \(3ab\times5b=15ab^2\).
   4. HCF is \(4x^2\): \(4x^2(5x-2)\) The HCF of 20 and 8 is 4, and the lowest power of \(x\) is \(x^2\). The factorisation is complete because \(5x-2\) has no common factor.

6. Answer to Question 6

   1. Two numbers with product 12 and sum 7: 3 and 4. \((x+3)(x+4)\) Check: \(3+4=7\) and \(3\times4=12\), so \(x^2+7x+12=(x+3)(x+4)\).
   2. Product \(-20\), sum \(-1\): \(-5\) and 4. \((x-5)(x+4)\) Check: \(-5+4=-1\) and \(-5\times4=-20\), so \(x^2-x-20=(x-5)(x+4)\).
   3. Product 21, sum \(-10\): \(-3\) and \(-7\). \((x-3)(x-7)\) A positive product with a negative sum means both numbers are negative, so \(x^2-10x+21=(x-3)(x-7)\).
   4. \(x^2-7^2=\) \((x-7)(x+7)\) There is no \(x\) term and 49 is a square number, so we use \(a^2-b^2=(a-b)(a+b)\).

7. Answer to Question 7

   1. \(4\) miles: \(3.50+7.20=\) £10.70. \(9.5\) miles: \(3.50+17.10=\) £20.60. \(14\) miles: \(3.50+25.20=\) £28.70. In each case multiply the miles by £1.80, then add the fixed charge of £3.50.
   2. £3.50 is the fixed starting charge and £1.80 is the price for each mile. For example, a journey of 0 miles would still cost £3.50 and each extra mile adds £1.80.
   3. \(21.50=3.50+1.80m\), so \(1.80m=18\) and \(m=\) 10 miles. Subtract the fixed charge first, then divide by the price per mile: \(18\div1.80=10\). Check: \(3.50+18=21.50\).
   4. \(3.50+1.80m=2+2m\) gives \(1.50=0.2m\), so \(m=7.5\). The firms charge the same for 7.5 miles (£17). Check with the other firm: \(2+2\times7.5=17\), and \(3.50+1.80\times7.5=17\). Both firms charge £17.

8. Answer to Question 8

   1. \(P=2(2x+3)+2(x-1)=4x+6+2x-2=\) \(6x+4\) metres. Opposite sides are equal, so there are two lengths and two widths in the perimeter.
   2. \(A=(2x+3)(x-1)=2x^2-2x+3x-3=\) \(2x^2+x-3\) m². Each term in the first bracket multiplies each term in the second, then \(-2x+3x=x\).
   3. When \(x=5\): \(P=34\) m and \(A=50+5-3=\) 52 m². (Check: length 13, width 4, \(13\times4=52\).) Substituting into \(6x+4\): \(6\times5+4=34\). So the perimeter is 34 m and the area is 52 m².
   4. \(34\times6=\) £204. The fencing goes round the perimeter, so the total cost of the fencing is £204.

9. Answer to Question 9

   1. \(B=12+0.05t+3d\) Texts cost 5p, which is £0.05, so every amount in the formula is in pounds.
   2. \(B=12+0.05\times40+3\times2=12+2+6=\) £20 Mo's bill for that month is therefore £20.
   3. \(22=12+0.05t+6\), so \(0.05t=4\) and \(t=4\div0.05=\) 80 texts. The fixed parts are \(12+3\times2=18\), so the texts cost \(22-18=4\) pounds. Mo sent 80 extra texts.
   4. \(B-12-3d=0.05t\), so \(t=\dfrac{B-12-3d}{0.05}\) (or \(20(B-12-3d)\)). Subtract \(12\) and \(3d\) from both sides, then divide by 0.05. Dividing by 0.05 is the same as multiplying by 20.

10. Answer to Question 10

    1. \(7+3=10\), \(10\times4=40\), \(40-12=\) 28. Follow the steps in order, so Ben ends up with 28.
    2. \(4(n+3)-12=4n+12-12=\) \(4n\). We put brackets around \(n+3\) because the whole result of the first step is multiplied by 4.
    3. The result is always 4 times the starting number. The \(+12\) from the expansion cancels the \(-12\) at the end. For example, starting with 7 gives \(4\times7=28\), which matches part (a).
    4. \(3(n+5)-9=3n+15-9=\) \(3n+6\). Expand the bracket first, then collect the numbers: \(15-9=6\). The result is \(3n+6\).

11. Answer to Question 11

    1. \((x+5)^2=x^2+5x+5x+25=\) \(x^2+10x+25\) m². The square has side \(x+5\), so its area is \((x+5)(x+5)\); remember the two middle terms.
    2. \(x^2+10x+25-x^2=\) \(10x+25\) m². The flower bed has area \(x^2\), which is subtracted from the large square, so the area to be paved is \(10x+25\) m².
    3. \((x+5)^2-x^2=(x+5-x)(x+5+x)=5(2x+5)\), which is \(5(2x+5)\) (the same as \(10x+25\)). Here \(a=x+5\) and \(b=x\). Expanding \(5(2x+5)\) gives back \(10x+25\), which checks the factorisation.
    4. \(5(6+5)=\) 55 m². When \(x=3\), \(2x+5=11\), so the area to be paved is \(5\times11=55\) m². Check: \(10\times3+25=55\).

12. Answer to Question 12

    1. Friday: \(35a+52c\). Saturday: \(48a+39c\). Each adult ticket brings in £\(a\) and each child ticket brings in £\(c\), so we multiply by the number of tickets sold.
    2. \(35a+48a+52c+39c=\) \(83a+91c\) Collect the \(a\) terms and the \(c\) terms separately because they are unlike terms.
    3. \(83\times6+91\times3.5=498+318.50=\) £816.50 Substitute \(a=6\) and \(c=3.5\) into \(83a+91c\). The total money taken over the two days is £816.50.
    4. No. The numbers 83 and 91 have no common factor greater than 1 (\(83\) is prime and \(91=7\times13\)), so the expression cannot be factorised. To factorise, the two coefficients would need a common factor, and they do not.

13. Answer to Question 13

    1. \(V=x^2(x+4)=\) \(x^3+4x^2\) cm³. The volume of a cuboid is base area times height, and the base area is \(x\times x=x^2\).
    2. \(2x^2+4x^2+16x=\) \(6x^2+16x\) cm². Multiply out \(4x(x+4)=4x^2+16x\), then collect \(2x^2+4x^2=6x^2\).
    3. The HCF of \(6x^2\) and \(16x\) is \(2x\): \(2x(3x+8)\). Check by expanding: \(2x\times3x=6x^2\) and \(2x\times8=16x\).
    4. Surface area \(=2\times6\times(18+8)=312\) cm². Volume \(=216+144=360\) cm³. (Check: \(6\times6\times10=360\).) Using \(2x(3x+8)\) with \(x=6\): \(12\times26=312\). So the surface area is 312 cm² and the volume is 360 cm³.

14. Answer to Question 14

    1. Zara: \(10x+15\). Leo: \(10(x+2)=\) \(10x+20\). Leo saves £\((x+2)\) for 10 weeks, so the whole bracket is multiplied by 10.
    2. Leo − Zara \(=10x+20-(10x+15)=\) £5. The difference does not depend on \(x\). The brackets are important: the minus sign changes the sign of both \(10x\) and \(15\). So Leo always has £5 more than Zara.
    3. Leo saves £6, so \(x+2=6\) and \(x=4\). Zara: \(10\times4+15=\) £55. Check: Leo has \(10\times6=60\) pounds, which is £5 more than Zara's £55.

15. Answer to Question 15

    1. \(n+(n+1)+(n+2)=3n+3=3(n+1)\), which is a multiple of 3 for every whole number \(n\). Since \(n+1\) is a whole number, \(3(n+1)\) is 3 times a whole number. So the sum of three consecutive whole numbers is always a multiple of 3.
    2. Of two consecutive numbers, exactly one is even, and an even number times any whole number is even, so \(n(n+1)\) is always even. If \(n\) is even, the product is even; if \(n\) is odd, then \(n+1\) is even, so the product is still even.
    3. \((n+1)^2-n^2=n^2+2n+1-n^2=2n+1\), which is odd (2 times a whole number plus 1). For example, \(5^2-4^2=25-16=9=2\times4+1\). Since \(2n\) is always even, \(2n+1\) is always odd.
    4. \(2n+1=51\) gives \(n=25\), so \(26^2-25^2=676-625=51\). So \(51=26^2-25^2\), which is the difference of two consecutive squares.

16. Answer to Question 16

    1. The side is \(x+4+4=\) \(x+8\) cm. The border adds 4 cm on the left and 4 cm on the right of the photo.
    2. \((x+8)^2-x^2=x^2+16x+64-x^2=\) \(16x+64\). The area of the border is the area of the whole frame minus the area of the photo.
    3. \(16(x+4)\) The HCF of 16 and 64 is 16. Check: \(16\times x+16\times4=16x+64\).
    4. \(16x+64=144\), so \(16x=80\) and \(x=5\). The photo is \(5\times5=\) 25 cm². Check: \((5+8)^2-5^2=169-25=144\). So \(x=5\) and the photo has area 25 cm².

17. Answer to Question 17

    1. \((x-9)(x+9)\), \((2x-5)(2x+5)\), \((3a-4b)(3a+4b)\). In each case write both terms as squares: \(81=9^2\), \(4x^2=(2x)^2\), \(25=5^2\), \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\).
    2. \(26\times100=\) 2600. Here \(63-37=26\) and \(63+37=100\), so \(63^2-37^2=2600\). This is much quicker than squaring each number.
    3. \((1.02-0.98)(1.02+0.98)=0.04\times2=\) 0.08. Here \(1.02-0.98=0.04\) and \(1.02+0.98=2\), so \(1.02^2-0.98^2=0.08\).
    4. \(x^4-1=(x^2-1)(x^2+1)=\) \((x-1)(x+1)(x^2+1)\). First treat \(x^4\) as \((x^2)^2\). Then \(x^2-1\) is also a difference of two squares, but \(x^2+1\) cannot be factorised.

18. Answer to Question 18

    1. \(ac=6\), numbers 1 and 6: \(2x^2+x+6x+3=x(2x+1)+3(2x+1)=\) \((2x+1)(x+3)\) Check by expanding: \(2x^2+6x+x+3=2x^2+7x+3\).
    2. \(ac=24\), numbers \(-4\) and \(-6\): \(3x^2-4x-6x+8=x(3x-4)-2(3x-4)=\) \((3x-4)(x-2)\) The product is positive and the sum is negative, so both numbers are negative. Check: \(3x^2-6x-4x+8=3x^2-10x+8\).
    3. \(ac=-72\), numbers 9 and \(-8\): \(6x^2+9x-8x-12=3x(2x+3)-4(2x+3)=\) \((2x+3)(3x-4)\) Check by expanding: \(6x^2-8x+9x-12=6x^2+x-12\).
    4. \((2x+1)(x+3)=0\), so \(x=-\tfrac12\) or \(x=-3\). If a product is zero, one of the brackets is zero: \(2x+1=0\) gives \(x=-\tfrac12\) and \(x+3=0\) gives \(x=-3\).

19. Answer to Question 19

    1. \(\dfrac{(x+2)(x+3)}{(x-2)(x+2)}=\) \(\dfrac{x+3}{x-2}\) The common factor \((x+2)\) cancels, so the fraction simplifies to \(\dfrac{x+3}{x-2}\).
    2. \(\dfrac{2x(x+3)}{(x-3)(x+3)}=\) \(\dfrac{2x}{x-3}\) The top has common factor \(2x\) and the bottom is a difference of two squares; \((x+3)\) cancels.
    3. \(\dfrac{(x-4)(x+3)}{(x+3)(x-2)}=\) \(\dfrac{x-4}{x-2}\) Both quadratics contain the factor \((x+3)\), which cancels top and bottom.
    4. \(\dfrac{3(x-2)+2(x+1)}{(x+1)(x-2)}=\) \(\dfrac{5x-4}{(x+1)(x-2)}\) The common denominator is \((x+1)(x-2)\). The numerator is \(3x-6+2x+2=5x-4\).

20. Answer to Question 20

    1. Row 1: \(x+3+x-4+x+1=3x\). Row 2: \(x-2+x+x+2=3x\). Row 3: \(x-1+x+4+x-3=3x\). Every row adds up to \(3x\) because the numbers in each row cancel out: \(3-4+1=0\), \(-2+0+2=0\) and \(-1+4-3=0\).
    2. Column 1: \(x+3+x-2+x-1=3x\). Diagonal: \(x+3+x+x-3=3x\). The column and the diagonal also add to \(3x\), so the square is magic.
    3. Middle: \(2x+5\). Magic sum: \(3(2x+5)=\) \(6x+15\). The middle entry \(x\) becomes \(2\times x+5=2x+5\), and the magic sum is three times the middle entry.
    4. Each of the 3 numbers in a line is doubled and has 5 added, so the sum is \(2\times3x+3\times5=6x+15\). So the new square is still magic, and its magic sum is \(6x+15\).