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

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1. Answer to Question 1

   1. \(3\times x+3\times5=\) \(3x+15\) Each of the two terms in the bracket is multiplied by 3, so the answer is \(3x+15\).
   2. \(4\times2y-4\times3=\) \(8y-12\) Both terms are multiplied by 4; keep the minus sign with the 12, giving \(8y-12\).
   3. \(-2\times a+(-2)\times6=\) \(-2a-12\) The negative number outside multiplies both terms, and a negative times a positive is negative, so the answer is \(-2a-12\).
   4. \(x\times x+x\times7=\) \(x^2+7x\) Multiplying \(x\) by \(x\) gives \(x^2\) and \(x\) by 7 gives \(7x\), so the answer is \(x^2+7x\).

2. Answer to Question 2

   1. \(3x+6+2x+10=\) \(5x+16\) Collect the \(x\) terms \(3x+2x=5x\) and the numbers \(6+10=16\).
   2. \(10a-5-3a-12=\) \(7a-17\) Collect \(10a-3a=7a\) and \(-5-12=-17\); the second bracket is multiplied by \(-3\), so both of its terms become negative.
   3. \(4y+12-y+2=\) \(3y+14\) (the minus changes both signs in the bracket) The bracket \(-(y-2)\) means \(-1\times(y-2)\), which gives \(-y+2\). Then \(4y-y=3y\) and \(12+2=14\).
   4. \(x^2+3x+2x+2=\) \(x^2+5x+2\) Only the \(x\) terms \(3x\) and \(2x\) can be collected; \(x^2\) and the number 2 stay as they are.

3. Answer to Question 3

   1. \(x^2+5x+3x+15=\) \(x^2+8x+15\) The four products are \(x^2\), \(5x\), \(3x\) and 15; the two middle terms add to \(8x\).
   2. \(x^2-2x+4x-8=\) \(x^2+2x-8\) The middle terms are \(-2x\) and \(+4x\), which add to \(+2x\); the last product \(4\times(-2)\) is \(-8\).
   3. \(x^2-6x-3x+18=\) \(x^2-9x+18\) Both numbers are negative, so the middle terms add to \(-9x\) and the constant \((-3)\times(-6)=+18\) is positive.
   4. \(2x^2+8x+x+4=\) \(2x^2+9x+4\) The first product is \(2x\times x=2x^2\); the middle terms \(8x\) and \(x\) give \(9x\), and the constant is \(1\times4=4\).

4. Answer to Question 4

   1. \(x^2+4x+4x+16=\) \(x^2+8x+16\) Write the bracket twice: the middle terms \(4x+4x\) give \(8x\) and the constant is \(4^2=16\).
   2. \(x^2-5x-5x+25=\) \(x^2-10x+25\) The middle terms are both \(-5x\), giving \(-10x\), and \((-5)^2=+25\).
   3. \(x^2-6x+6x-36=\) \(x^2-36\) (the middle terms cancel) The brackets differ only in their sign, so \(-6x\) and \(+6x\) cancel and the answer is \(x^2-6^2\).
   4. \(9x^2+6x+6x+4=\) \(9x^2+12x+4\) Here \((3x)^2=9x^2\), the middle terms are \(2\times3x\times2=12x\) and the constant is \(2^2=4\).

5. Answer to Question 5

   1. HCF is 3: \(3(2x+3)\) Dividing each term by 3 leaves \(2x\) and 3. Check: \(3\times2x+3\times3=6x+9\).
   2. HCF is 4: \(4(2y-3)\) Dividing by 4 leaves \(2y\) and \(-3\). Check: \(4\times2y-4\times3=8y-12\).
   3. HCF is \(x\): \(x(x+5)\) There is no common number other than 1, but \(x\) divides both terms, leaving \(x\) and 5 inside the bracket.
   4. HCF is \(6a\): \(6a(2b+3)\) The HCF of 12 and 18 is 6, and \(a\) appears in both terms but \(b\) does not, so the HCF is \(6a\). Check: \(6a\times2b+6a\times3=12ab+18a\).

6. Answer to Question 6

   1. No. \(5(x+2)=5x+10\), so the claim is wrong. The 5 must multiply both terms in the bracket, not only the \(x\), so the correct expansion ends in \(+10\).
   2. \(2x(x+3)\). Check: \(2x\times x+2x\times3=2x^2+6x\). Correct. The HCF of \(2x^2\) and \(6x\) is \(2x\), because 2 divides both numbers and the lowest power of \(x\) is \(x\).
   3. \(x^2-49=x^2-7^2=\) \((x+7)(x-7)\) Since \(49=7^2\), this is a difference of two squares \(a^2-b^2\) with \(a=x\) and \(b=7\).
   4. \(4x^2-9=(2x)^2-3^2=\) \((2x+3)(2x-3)\) Since \(4x^2=(2x)^2\) and \(9=3^2\), use \(a^2-b^2=(a+b)(a-b)\) with \(a=2x\) and \(b=3\). Check: \(4x^2-6x+6x-9=4x^2-9\).

7. Answer to Question 7

   1. \((x+6)(x+2)=x^2+2x+6x+12=\) \(x^2+8x+12\) Multiply each term of the first bracket by each term of the second, then collect \(2x+6x=8x\). This shows the area is \(x^2+8x+12\) m².
   2. \(16+32+12=60\). Check: \(10\times6=\) 60 m². When \(x=4\) the sides are 10 m and 6 m. The expression and the actual lengths give the same value, which confirms the expansion: the area is 60 m².
   3. \(2(x+6)+2(x+2)=2x+12+2x+4=\) \(4x+16\) A rectangle has two lengths and two widths, so add twice each side and collect the like terms: the perimeter is \((4x+16)\) m.
   4. When \(x=4\) the perimeter is \(16+16=32\) m. Cost \(=32\times8.50=\) £272 First find the perimeter, then multiply by the price per metre. The fencing costs £272.

8. Answer to Question 8

   1. \((2x+1)(x+4)=2x^2+8x+x+4=\) \(2x^2+9x+4\) Multiply out the four products \(2x^2\), \(8x\), \(x\) and 4, then collect \(8x+x=9x\). This proves the result.
   2. Expression: \(50+45+4=99\). Lengths: \(11\times9=\) 99 m². They agree. When \(x=5\) the sides are \(2\times5+1=11\) m and \(5+4=9\) m, so both methods give an area of 99 m².
   3. \(99\div20=4.95\), so 5 tins. Four tins only cover 80 m², which is not enough, so we must round up: the caretaker needs 5 tins.
   4. \(5\times14.20=\) £71 Multiply the number of tins by the price of one tin: the total cost is £71.

9. Answer to Question 9

   1. Side \(=x+3+3=\) \(x+6\) There is a 3 cm border on the left and on the right of the picture, so 6 cm is added to the side \(x\).
   2. \((x+6)^2=x^2+6x+6x+36=\) \(x^2+12x+36\) The frame is a square, so its area is \((x+6)(x+6)\); the two middle terms \(6x\) add to \(12x\) and \(6^2=36\).
   3. Border area \(=(x^2+12x+36)-x^2=\) \(12x+36\) The border is the whole frame without the picture. The picture has area \(x^2\), which cancels, leaving \(12x+36\) cm².
   4. \(12\times20+36=276\). Check: \(26^2-20^2=676-400=\) 276 cm². The frame has side 26 cm and the picture has side 20 cm, so both methods agree: the border is 276 cm².

10. Answer to Question 10

    1. \(2(x+2)+3x=2x+4+3x=\) \(5x+4\) Each adult ticket costs \(x+2\) pounds, so two of them cost \(2(x+2)\); add \(3x\) for the children's tickets. The total is \(£(5x+4)\).
    2. \(5\times3+4=\) £19 Substitute \(x=3\): \(15+4=19\). The Brown family pays £19.
    3. \(4(x+2)+6x=4x+8+6x=10x+8=\) \(2(5x+4)\) The HCF of 10 and 8 is 2, so \(10x+8=2(5x+4)\). The Greens pay \(£2(5x+4)\).
    4. The Green family buys exactly twice as many tickets, so they pay exactly twice as much as the Browns. The bracket \(5x+4\) is exactly the Browns' total, and it is multiplied by 2, so the Greens pay double for any ticket price.

11. Answer to Question 11

    1. \((100+3)(100-3)=100^2-3^2=10000-9=\) 9 991 Here \(a=100\) and \(b=3\), so \(a^2-b^2\) is easy to work out mentally. The answer is 9 991.
    2. \(50^2+2\times50\times2+2^2=2500+200+4=\) 2 704 Use \((a+b)^2=a^2+2ab+b^2\) with \(a=50\) and \(b=2\). So \(52^2=2704\).
    3. \((50-2)(50+2)=2500-4=\) 2 496 Write the product as \((50-2)(50+2)\), a difference of two squares, so it equals \(50^2-2^2=2496\).
    4. \((100-1)^2=10000-200+1=\) 9 801 Use \((a-b)^2=a^2-2ab+b^2\) with \(a=100\) and \(b=1\). So \(99^2=9801\).

12. Answer to Question 12

    1. \(x+3\) Each box contains \(x\) muffins and 3 cookies, so one box holds \(x+3\) items.
    2. \(4(x+3)+6(x+3)=4x+12+6x+18=\) \(10x+30\) There are \(4+6=10\) boxes, each with \(x+3\) items. Expanding and collecting gives \(10x+30\) items.
    3. \(10x+30=\) \(10(x+3)\), which is 10 boxes of \(x+3\) items. The HCF of 10 and 30 is 10, so the total factorises as \(10(x+3)\): ten boxes of \(x+3\) items.
    4. \(10\times(9+3)=\) 120 items Use the factorised form: \(9+3=12\) items per box and 10 boxes, so the baker packs 120 items in a day.

13. Answer to Question 13

    1. \(x^2+3x+2x+6=\) \(x^2+5x+6\) The four products are \(x^2\), \(3x\), \(2x\) and 6; collecting the middle terms gives \(5x\).
    2. \(x(x^2+5x+6)=\) \(x^3+5x^2+6x\) Volume = length × width × height, so multiply the answer to part a) by \(x\). Each term gains one power of \(x\).
    3. Expression: \(64+80+24=168\). Lengths: \(4\times6\times7=\) 168 cm³. Both methods give the same value, so the volume is 168 cm³ when \(x=4\).
    4. \(188\times0.2=37.6\) p, so about 38p. Multiply the surface area by the cost per cm². Check the surface area: \(2(4\times6+4\times7+6\times7)=2\times94=188\) cm². The card costs about 38p.

14. Answer to Question 14

    1. \(4(2x+5)=\) \(8x+20\) Jade saves the same amount every week, so multiply by 4: her total is \(£(8x+20)\).
    2. \(3(x+8)=\) \(3x+24\) Ben saves for 3 weeks, so his total is \(£(3x+24)\).
    3. \(8x+20+3x+24=11x+44=\) \(11(x+4)\) Add the two totals and collect like terms: \(11x+44\). The HCF of 11 and 44 is 11, so the total is \(£11(x+4)\).
    4. \(11\times(6+4)=\) £110 Substitute \(x=6\) into the factorised form: \(11\times10=110\). Together they save £110.

15. Answer to Question 15

    1. \((x-2)(x+3)=x^2+3x-2x-6=\) \(x^2+x-6\). \((x+2)(x-3)=\) \(x^2-x-6\). \((x-2)(x-3)=\) \(x^2-5x+6\). In each case multiply the four pairs of terms and collect the two \(x\) terms; take care that a negative times a positive is negative.
    2. The coefficient of \(x\) is the sum of the two numbers and the constant is their product. For \((x+a)(x+b)\) the middle terms are \(bx+ax=(a+b)x\) and the last term is \(ab\), which explains the pattern.
    3. Sum 11, product 28: \(x^2+11x+28\) The numbers are 4 and 7: their sum is 11 and their product is 28, so the expansion is \(x^2+11x+28\).
    4. Sum \(=3\), product \(=-40\): \(x^2+3x-40\). Check: \(x^2+8x-5x-40=\) \(x^2+3x-40\). Correct. The numbers are \(-5\) and 8: \(-5+8=3\) and \(-5\times8=-40\). The prediction and the full expansion agree.

16. Answer to Question 16

    1. Ali wrong, Bea wrong, Cal right, Dev wrong. Expand each one yourself and compare. Only Cal has every term correct; each of the others has a missing term or a wrong sign.
    2. Ali: \((x+3)(x+3)=x^2+6x+9\). Bea: \(3x-12\). Dev: \(-3x+6\) (a negative times a negative is positive). Ali forgot the middle term, Bea forgot to multiply the 4 by 3, and Dev forgot that \(-3\times(-2)=+6\).
    3. When \(x=1\): \((1+3)^2=16\) but \(1+9=10\). The two expressions are not equivalent: the middle term \(6x\) was forgotten. Substituting a number shows the two sides have different values, so \((x+3)^2\) and \(x^2+9\) cannot be equivalent.
    4. Cal's: \((x-2)(x+2)=x^2-4\). Cal's working has the form \((a-b)(a+b)=a^2-b^2\) with \(a=x\) and \(b=2\), so it is the difference of two squares.

17. Answer to Question 17

    1. HCF is \(3x\): \(6x^2+9x=\) \(3x(2x+3)\) The HCF of 6 and 9 is 3 and both terms contain \(x\), so take out \(3x\). Check: \(3x\times2x+3x\times3=6x^2+9x\).
    2. Other side \(=\) \((2x+3)\) m Area = length × width, and one side is \(3x\), so the other side is the bracket left after factorising.
    3. \(2(3x+2x+3)=\) \(10x+6\) metres Add the two different sides and double: \(3x+2x+3=5x+3\), so the perimeter is \((10x+6)\) m.
    4. Sides \(3x=6\) m and \(2x+3=7\) m. Area \(=42\) m\(^2\). Check: \(6\times4+18=24+18=42\). Substituting into the expression and multiplying the side lengths both give 42 m², so the patio is 6 m by 7 m.
    5. \(42\times7.50=\) £315 Multiply the area by the price per square metre: the slabs cost £315.

18. Answer to Question 18

    1. \(n=2\): \((2+4)^2=36\), \(36-16=20\), \(20\div2=10\). \(n=5\): \(81-16=65\), \(65\div5=13\). \(n=10\) also works: \(14^2=196\), \(196-16=180\) and \(180\div10=18\). All the entries in the table are correct.
    2. \(\dfrac{(n+4)^2-16}{n}\) Follow the steps in order: adding 4 gives \(n+4\), squaring gives \((n+4)^2\), subtract 16, then divide the whole result by \(n\).
    3. \((n+4)^2-16=n^2+8n+16-16=n^2+8n=n(n+8)\) The numbers 16 and \(-16\) cancel, and \(n\) is a common factor of \(n^2\) and \(8n\), giving \(n(n+8)\).
    4. \(\dfrac{n(n+8)}{n}=\) \(n+8\) for every \(n\ne0\). The factor \(n\) on the top cancels with the \(n\) on the bottom, so the final answer is always 8 more than the starting number.
    5. Dividing by zero is impossible, so \(n=0\) is not allowed. When \(n=0\) the last step asks us to divide 0 by 0, which has no value, so the trick only works for non-zero numbers.

19. Answer to Question 19

    1. Width \(x+4\), height \(x+8\) (2 cm is added on each side). A 2 cm border on both sides adds \(2+2=4\) cm to the width and 4 cm to the height, so the poster is \((x+4)\) cm by \((x+8)\) cm.
    2. \((x+4)(x+8)=x^2+8x+4x+32=\) \(x^2+12x+32\) Multiply the four products \(x^2\), \(8x\), \(4x\) and 32, then collect \(8x+4x=12x\).
    3. Picture area \(=x(x+4)=x^2+4x\). Border \(=x^2+12x+32-x^2-4x=\) \(8x+32=8(x+4)\) Subtract the picture area from the poster area; the \(x^2\) terms cancel and \(12x-4x=8x\). The HCF of 8 and 32 is 8.
    4. \(8\times16=128\). Check: \(16\times20-12\times16=320-192=\) 128 cm². When \(x=12\) the poster is 16 cm by 20 cm and the picture is 12 cm by 16 cm, so the border area is 128 cm² both ways.
    5. \(128\times0.5=\) 64p Multiply the border area by the cost per cm²: the paper costs 64p.

20. Answer to Question 20

    1. \((2k+3)^2=4k^2+12k+9\) and \((2k+1)^2=4k^2+4k+1\). Use \((a+b)^2=a^2+2ab+b^2\): for \(2k+3\) the middle term is \(2\times2k\times3=12k\), and for \(2k+1\) it is \(4k\).
    2. Subtract: \((4k^2+12k+9)-(4k^2+4k+1)=8k+8\). The \(4k^2\) terms cancel, \(12k-4k=8k\) and \(9-1=8\), which proves the result.
    3. \(8k+8=8(k+1)\), which is a multiple of 8. Since 8 is a factor of \(8(k+1)\) and \(k+1\) is a whole number, the difference is always a multiple of 8.
    4. \(49-25=24=8\times3\). (Here \(k=2\), so \(8(k+1)=24\).) The odd numbers 5 and 7 give \(2k+1=5\), so \(k=2\), and the formula predicts \(8\times3=24\), which matches.
    5. \((51-49)(51+49)=2\times100=200=8\times25\). Using \(a^2-b^2=(a-b)(a+b)\) avoids squaring large numbers. Since \(200=8\times25\), it is a multiple of 8.