Expanding Brackets

Multiply every term in the bracket by 2x^3, adding the powers of x.

1. 2x^3\times4x^2=8x^5
2. 2x^3\times(-3x)=-6x^4
3. 2x^3\times5=10x^3
4. Answer: 8x^5-6x^4+10x^3

Write the square as (4x-3)(4x-3) so that you don't miss the middle terms.

1. (4x-3)^2=(4x-3)(4x-3)
2. =16x^2-12x-12x+9
3. Not 16x^2+9: squaring each term separately misses the two middle terms
4. Answer: 16x^2-24x+9

Multiply each of the three terms in the first bracket by x and then by 2; most of the terms cancel.

1. Multiply by x: x^3-2x^2+4x
2. Multiply by 2: 2x^2-4x+8
3. Add: x^3-2x^2+2x^2+4x-4x+8
4. The x^2 terms cancel and so do the x terms
5. Answer: x^3+8

Multiply every term in the second bracket by -4, so -4\times(-3)=+12.

1. 2x(3x-5)=6x^2-10x
2. -4(x^2-2x-3)=-4x^2+8x+12
3. Collect like terms: 6x^2-4x^2-10x+8x+12=2x^2-2x+12
4. Take out the common factor 2
5. Answer: 2(x^2-x+6)

You don't need the whole expansion: find every pair of terms, one from each bracket, that multiplies to give an x^2 term.

1. Look for the products that give x^2:
2. x^2\times 2=2x^2
3. -4x\times 3x=-12x^2
4. No other pair gives x^2 (7\times 3x gives an x term, x^2\times 3x gives x^3).
5. Add them: 2x^2-12x^2=-10x^2
6. The coefficient is the number in front of x^2: -10

Find every pair of terms, one from each bracket, that multiplies to give an xy term.

1. 3x\times4y=12xy
2. -2y\times x=-2xy
3. No other pair gives xy
4. 12xy-2xy=10xy
5. Answer: 10

Work out (x-4)^2 first, then multiply your answer by (x-4).

1. (x-4)^2=(x-4)(x-4)=x^2-8x+16
2. (x-4)^3=(x^2-8x+16)(x-4)
3. =x^3-8x^2+16x-4x^2+32x-64
4. Answer: x^3-12x^2+48x-64

Multiply two of the brackets together first and simplify, then multiply your answer by the third bracket.

1. Expand the first two brackets: (2x+5)(x-4)=2x^2-8x+5x-20=2x^2-3x-20
2. Now multiply by (3x-1): (2x^2-3x-20)(3x-1)
3. 2x^2\times 3x=6x^3, \ 2x^2\times(-1)=-2x^2
4. -3x\times 3x=-9x^2, \ -3x\times(-1)=+3x
5. -20\times 3x=-60x, \ -20\times(-1)=+20
6. Collect like terms: 6x^3-11x^2-57x+20
7. Answer: 6x^3-11x^2-57x+20

Write it as \left(\sqrt{x}+\dfrac{3}{\sqrt{x}}\right)\left(\sqrt{x}+\dfrac{3}{\sqrt{x}}\right) and use \sqrt{x}\times\sqrt{x}=x.

1. \sqrt{x}\times\sqrt{x}=x
2. \sqrt{x}\times\dfrac{3}{\sqrt{x}}=3, and this term appears twice
3. \dfrac{3}{\sqrt{x}}\times\dfrac{3}{\sqrt{x}}=\dfrac{9}{x}
4. x+3+3+\dfrac{9}{x}
5. Answer: x+6+\dfrac{9}{x}

Find the two products that give an x term; one of them involves k.

1. x terms: x\times5=5x and k\times(-3x)=-3kx
2. So the coefficient of x is 5-3k
3. 5-3k=-7
4. -3k=-12
5. Answer: k=4

Replace p and q with their brackets, and put a bracket round 2pq before you subtract it.

1. p^2=(2x-1)^2=4x^2-4x+1
2. 2pq=2(2x-1)(x+3)=2(2x^2+5x-3)=4x^2+10x-6
3. p^2-2pq=4x^2-4x+1-(4x^2+10x-6)
4. =4x^2-4x+1-4x^2-10x+6
5. Answer: 7-14x

Expand (3-2x)(x+4) first and simplify, then multiply by (x-1).

1. (3-2x)(x+4)=3x+12-2x^2-8x=-2x^2-5x+12
2. (-2x^2-5x+12)(x-1)=-2x^3+2x^2-5x^2+5x+12x-12
3. =-2x^3-3x^2+17x-12
4. The x^3 term is negative because -2x\times x\times x=-2x^3: don't change the signs to make it positive
5. Answer: -2x^3-3x^2+17x-12

Expand each part separately, remembering (2x-3)^2=(2x-3)(2x-3), then add the two results.

1. (x^2+2)(4x-1)=4x^3-x^2+8x-2
2. (2x-3)^2=4x^2-12x+9
3. x(4x^2-12x+9)=4x^3-12x^2+9x
4. Add the two parts (don't multiply them): 4x^3-x^2+8x-2+4x^3-12x^2+9x
5. Answer: 8x^3-13x^2+17x-2

The lawn and path together make a bigger rectangle, 4 m longer and 4 m wider than the lawn.

1. The outer rectangle is (2x+5) m by (x+7) m
2. Outer area: (2x+5)(x+7)=2x^2+19x+35
3. Lawn area: (2x+1)(x+3)=2x^2+7x+3
4. Path =2x^2+19x+35-(2x^2+7x+3)
5. Answer: 12x+32 m^2

Expand both sides fully: the x^2 terms cancel and leave a linear inequality.

1. Left: (x+3)(2x-5)=2x^2+x-15
2. Right: 2(x-1)^2=2(x^2-2x+1)=2x^2-4x+2
3. 2x^2+x-15\geqslant2x^2-4x+2
4. Subtract 2x^2 from both sides: x-15\geqslant-4x+2
5. 5x\geqslant17
6. Answer: x\geqslant\dfrac{17}{5}

Use Pythagoras with (2x-3) as the hypotenuse, writing each squared length as a pair of brackets before expanding. Answer part (a) before attempting part (b).

Part (a)

1. (x-3)^2+(x+4)^2=(2x-3)^2
2. Left: x^2-6x+9+x^2+8x+16=2x^2+2x+25
3. Right: 4x^2-12x+9
4. 2x^2+2x+25=4x^2-12x+9
5. 0=2x^2-14x-16
6. Divide by 2
7. Answer: x^2-7x-8=0

Part (b)

1. x^2-7x-8=(x-8)(x+1)=0
2. x=8 or x=-1
3. x=-1 would make the sides -4 cm, 3 cm and -5 cm, which is impossible
4. Answer: x=8

Part (c)

1. The sides are 5 cm, 12 cm and 13 cm
2. The two shorter sides are perpendicular
3. Area =\dfrac{1}{2}\times5\times12
4. Answer: 30 cm^2

Start with the constant terms: they only involve k. Then put your value of k back in and expand everything, remembering (x-3)^2=(x-3)(x-3).

Part (a)

1. Compare the constant terms (the terms with no x).
2. (x-3)^2=x^2-6x+9, so the constant in (2x+k)(x-3)^2 is k\times 9=9k
3. (x+2)^3 has constant term 2^3=8
4. So 9k-8=-26
5. 9k=-18
6. k=-2

Part (b)

1. Use k=-2: (2x-2)(x^2-6x+9)=2x^3-12x^2+18x-2x^2+12x-18
2. =2x^3-14x^2+30x-18
3. (x+2)^3=(x+2)(x^2+4x+4)=x^3+6x^2+12x+8
4. Subtract, using brackets: 2x^3-14x^2+30x-18-(x^3+6x^2+12x+8)
5. =x^3-20x^2+18x-26
6. Compare the x^2 terms: a=-20

Part (c)

1. From part (b): x^3-20x^2+18x-26
2. Compare the x terms: b=18

\mathrm{gf}(x) means put \mathrm{f}(x) into \mathrm{g}, so replace every x in \mathrm{g}(x) with (2x-3). Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{gf}(x)=\mathrm{g}(2x-3)
2. =(2x-3)^2+4(2x-3)
3. =4x^2-12x+9+8x-12
4. Answer: 4x^2-4x-3

Part (b)

1. \mathrm{fg}(x)=\mathrm{f}(x^2+4x)=2(x^2+4x)-3=2x^2+8x-3
2. 4x^2-4x-3=2x^2+8x-3
3. 2x^2-12x=0
4. 2x(x-6)=0
5. Answer: x=0 or x=6

Write (n+2)^3 as (n+2)(n+2)^2 and expand in two stages. Answer part (a) before attempting part (b).

Part (a)

1. (n+2)^2=n^2+4n+4
2. (n+2)^3=(n^2+4n+4)(n+2)
3. =n^3+2n^2+4n^2+8n+4n+8
4. =n^3+6n^2+12n+8
5. Subtract n^3
6. Answer: 6n^2+12n+8

Part (b)

1. 6n^2+12n+8=2(3n^2+6n+4) and 3n^2+6n+4 is an integer, so the expression is always even
2. 4(1.5n^2+3n+2) does not show a multiple of 4: the bracket need not be an integer
3. When n=1: 3^3-1^3=26, which is not a multiple of 4, 6 or 8
4. Answer: always even, because it equals 2(3n^2+6n+4)

Part (c)

1. 6n^2+12n+8=296
2. 6n^2+12n-288=0
3. Divide by 6: n^2+2n-48=0
4. (n+8)(n-6)=0
5. n is positive, so n\neq-8
6. Answer: n=6

You can only differentiate term by term once the brackets are expanded. Answer part (a) before attempting part (b).

Part (a)

1. (x-1)^2=x^2-2x+1
2. (x^2-2x+1)(4x+3)=4x^3+3x^2-8x^2-6x+4x+3
3. Answer: 4x^3-5x^2-2x+3

Part (b)

1. \dfrac{\mathrm{d}y}{\mathrm{d}x}=12x^2-10x-2
2. At x=2: 12\times4-10\times2-2
3. =48-20-2
4. Answer: 26

Part (c)

1. Stationary points: \dfrac{\mathrm{d}y}{\mathrm{d}x}=0
2. 12x^2-10x-2=0
3. Divide by 2: 6x^2-5x-1=0
4. (6x+1)(x-1)=0
5. Answer: x=-\dfrac{1}{6} and x=1