Year 9 Mathematics | Victorian Curriculum 2.0
Expanding and factorising
Topic 03 | Number & Algebra | Answer key

Tier 1

    1. x2+8x+12x^2 + 8x + 12.
    2. x2+2x−15x^2 + 2x - 15.
    3. x2−11x+28x^2 - 11x + 28.
    4. x2+16x+64x^2 + 16x + 64.
    5. x2−6x+9x^2 - 6x + 9.
    6. x2−121x^2 - 121.
    7. (x+3)(x+5)(x + 3)(x + 5). Method: 3+5=83 + 5 = 8 and 3×5=153 \times 5 = 15.
    8. (x−2)(x−3)(x - 2)(x - 3). Method: (−2)+(−3)=−5(-2) + (-3) = -5 and (−2)(−3)=6(-2)(-3) = 6.
    9. (x+6)(x−6)(x + 6)(x - 6). Difference of squares: 36=6236 = 6^2.
    10. (x+6)2(x + 6)^2. Perfect square: 12=2×612 = 2 \times 6 and 36=6236 = 6^2.

Tier 2

    1. (x2+7x+12)−(x2+3x+2)=4x+10(x^2 + 7x + 12) - (x^2 + 3x + 2) = 4x + 10. The x2x^2 terms cancel.
    2. (x+5)(x−4)(x + 5)(x - 4). Method: 5+(−4)=15 + (-4) = 1 and 5×(−4)=−205 \times (-4) = -20.
    3. (x−7)(x+4)(x - 7)(x + 4). Method: (−7)+4=−3(-7) + 4 = -3 and (−7)(4)=−28(-7)(4) = -28.
    4. Outer square side: (x+5+2)=(x+7)(x + 5 + 2) = (x + 7) m. Path area =(x+7)2−(x+5)2=(x2+14x+49)−(x2+10x+25)=4x+24= (x+7)^2 - (x+5)^2 = (x^2 + 14x + 49) - (x^2 + 10x + 25) = 4x + 24 m2^2.
    5. x2−100=(x+10)(x−10)x^2 - 100 = (x+10)(x-10). So 9982−10002=(998+1000)(998−1000)=1998×(−2)=−3996998^2 - 1000^2 = (998+1000)(998-1000) = 1998 \times (-2) = -3996.
    6. (x+a)2=x2+2ax+a2(x+a)^2 = x^2 + 2ax + a^2. (x−a)2=x2−2ax+a2(x-a)^2 = x^2 - 2ax + a^2. Subtracting: (x2+2ax+a2)−(x2−2ax+a2)=4ax(x^2 + 2ax + a^2) - (x^2 - 2ax + a^2) = 4ax.
    7. (x−7)2(x - 7)^2. Perfect square: 14=2×714 = 2 \times 7 and 49=7249 = 7^2.
    8. 2(x2+7x+12)=2(x+3)(x+4)2(x^2 + 7x + 12) = 2(x + 3)(x + 4). Factor out 22 first, then factorise the trinomial.

Tier 3

    1. When expanding (x+a)(x−a)(x+a)(x-a), the middle terms are +ax+ax and −ax-ax, which sum to zero. Using the area model: the two rectangular strips (a×xa \times x and x×ax \times a) have opposite signs and cancel, leaving only x2−a2x^2 - a^2.
    2. x2+9x+18=(x+3)(x+6)x^2 + 9x + 18 = (x + 3)(x + 6). Method: 3+6=93 + 6 = 9 and 3×6=183 \times 6 = 18. So the length is (x+6)(x + 6) cm and the width is (x+3)(x + 3) cm (or vice versa).
    3. They are not equivalent. (x+3)2=x2+6x+9(x+3)^2 = x^2 + 6x + 9, which has a 6x6x term that x2+9x^2 + 9 lacks. For example, at x=1x = 1: (1+3)2=16(1+3)^2 = 16 but 1+9=101 + 9 = 10.
    4. x2−2x−35=(x−7)(x+5)=0x^2 - 2x - 35 = (x - 7)(x + 5) = 0. So x=7x = 7 or x=−5x = -5.
    5. Expanding is like multiplication: 3×4=123 \times 4 = 12 breaks a product into a single value. Factorising is like finding factors: 12=3×412 = 3 \times 4 rewrites a value as a product. They undo each other. In algebra, expanding turns (x+3)(x+4)(x+3)(x+4) into x2+7x+12x^2 + 7x + 12, and factorising reverses the process.

Challenge

    1. (2x+3)(x−4)=2x2−8x+3x−12=2x2−5x−12(2x+3)(x-4) = 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12.
    2. x4−16=(x2)2−42=(x2+4)(x2−4)x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4). Then x2−4=(x+2)(x−2)x^2 - 4 = (x+2)(x-2), so x4−16=(x2+4)(x+2)(x−2)x^4 - 16 = (x^2 + 4)(x + 2)(x - 2). The factor (x2+4)(x^2 + 4) does not factorise further over the reals.
    3. Product of two consecutive odd numbers: (2n−1)(2n+1)=4n2−1(2n-1)(2n+1) = 4n^2 - 1. Their sum is (2n−1)+(2n+1)=4n(2n-1) + (2n+1) = 4n, which is divisible by 44. (Note: the question asks about the sum, not the product.)
    4. Square both sides of x+1x=5x + \dfrac{1}{x} = 5: x2+2+1x2=25x^2 + 2 + \dfrac{1}{x^2} = 25. So x2+1x2=25−2=23x^2 + \dfrac{1}{x^2} = 25 - 2 = 23.
Year 9 Mathematics study companion | Answer key