Binomial Expansion

The coefficients in the expansion of (a+b)^5 are the numbers in row 5 of Pascal's triangle.

1. Row 5 of Pascal's triangle is 1\;5\;10\;10\;5\;1
2. The terms in order are a^5,\ a^4b,\ a^3b^2,\ a^2b^3,\ ab^4,\ b^5
3. a^2b^3 is the fourth term, so its coefficient is the fourth number in the row
4. Answer: 10

In each term the powers of 4 and of 5x add up to 3, and the parts of a term are multiplied together.

1. Row 3 of Pascal's triangle is 1\;3\;3\;1
2. The x^2 term uses the third number, 3, with 4^1 and (5x)^2
3. The three parts are multiplied, and the whole of 5x is squared
4. 3\times4\times(5x)^2=3\times4\times25x^2=300x^2
5. (3\times4\times5x^2 squares only the x; 3\times4+(5x)^2 adds instead of multiplying; 4^2 makes the powers add up to 4.)
6. Answer: 3\times4\times(5x)^2

Use the row 1\;4\;6\;4\;1 of Pascal's triangle, and remember the whole term -3x is cubed.

1. Row 4 of Pascal's triangle is 1\;4\;6\;4\;1
2. The x^3 term uses the fourth number, 4, with 1^1 and (-3x)^3
3. (-3x)^3=-27x^3 (cube the -3 as well as the x)
4. 4\times 1\times(-27x^3)=-108x^3
5. Answer: -108x^3

Use the row 1\;3\;3\;1 of Pascal's triangle, and put brackets round 4x before you raise it to a power.

1. Row 3 of Pascal's triangle: 1\;3\;3\;1
2. 1\times(4x)^3=64x^3
3. 3\times(4x)^2\times(-1)=-48x^2
4. 3\times(4x)\times(-1)^2=12x
5. 1\times(-1)^3=-1
6. Answer: 64x^3-48x^2+12x-1

Use the third number in row 4 of Pascal's triangle, and square the whole of -3x.

1. Row 4 of Pascal's triangle: 1\;4\;6\;4\;1
2. The x^2 term is 6\times5^2\times(-3x)^2
3. (-3x)^2=9x^2
4. 6\times25\times9=1350
5. Answer: 1350

The term in x^2y^3 uses the fourth number in row 5 of Pascal's triangle, and the whole of -2y is cubed.

1. Row 5 of Pascal's triangle: 1\;5\;10\;10\;5\;1
2. The x^2y^3 term is 10\times x^2\times(-2y)^3
3. (-2y)^3=-8y^3
4. 10\times(-8)=-80
5. Answer: -80

Row 6 of Pascal's triangle starts 1\;6\;15; the powers of -4x go up from 0.

1. Row 6 of Pascal's triangle starts 1\;6\;15
2. First term: 1\times1^6=1
3. Second term: 6\times1^5\times(-4x)=-24x
4. Third term: 15\times1^4\times(-4x)^2=15\times16x^2=240x^2
5. Answer: 1-24x+240x^2

Square the whole term \dfrac{x}{3}, so the 3 on the bottom is squared too.

1. Row 5 of Pascal's triangle: 1\;5\;10\;10\;5\;1
2. The x^2 term is 10\times3^3\times\left(\dfrac{x}{3}\right)^2
3. \left(\dfrac{x}{3}\right)^2=\dfrac{x^2}{9}
4. 10\times27\times\dfrac19=30
5. Answer: 30

Expand both brackets using the row 1\;3\;3\;1; the terms with odd powers of 3 change sign in the second expansion.

1. (x+3)^3=x^3+9x^2+27x+27
2. (x-3)^3=x^3-9x^2+27x-27
3. Subtract: the x^3 and x terms cancel
4. 9x^2-(-9x^2)=18x^2 and 27-(-27)=54
5. Answer: 18x^2+54

Use the row 1\;3\;3\;1 with x^2 as the first term: (x^2)^3=x^6.

1. Row 3 of Pascal's triangle: 1\;3\;3\;1
2. 1\times(x^2)^3=x^6
3. 3\times(x^2)^2\times(-3)=-9x^4
4. 3\times x^2\times(-3)^2=27x^2
5. 1\times(-3)^3=-27
6. Answer: x^6-9x^4+27x^2-27

The x^2 term is (the third number in row n of Pascal's triangle) \times(3x)^2.

1. The x^2 term is (third number in row n) \times(3x)^2
2. (3x)^2=9x^2, so the third number in row n is 135\div9=15
3. Row 6 of Pascal's triangle is 1\;6\;15\;20\;15\;6\;1
4. Answer: n=6

Use the row 1\;4\;6\;4\;1 of Pascal's triangle, with powers of 2 going down and powers of 3x going up.

1. Row 4 of Pascal's triangle: 1\;4\;6\;4\;1
2. 1\times2^4=16
3. 4\times2^3\times(3x)=96x
4. 6\times2^2\times(3x)^2=6\times4\times9x^2=216x^2
5. 4\times2\times(3x)^3=4\times2\times27x^3=216x^3
6. 1\times(3x)^4=81x^4
7. Answer: 16+96x+216x^2+216x^3+81x^4

Write the x^3 coefficient in terms of k, remembering that (kx)^3=k^3x^3.

1. Row 4 of Pascal's triangle: 1\;4\;6\;4\;1
2. The x^3 term is 4\times3\times(kx)^3=12k^3x^3
3. 12k^3=-96
4. k^3=-8
5. A cube root has only one value: k=-2
6. Answer: k=-2

The x^3 term contains k^2, so there are two values of k.

1. Row 5 of Pascal's triangle: 1\;5\;10\;10\;5\;1
2. The x^3 term is 10\times k^2\times(-2x)^3
3. (-2x)^3=-8x^3, so the coefficient is -80k^2
4. -80k^2=-1280
5. k^2=16
6. Answer: k=4 or k=-4

Work out each coefficient separately (one will contain a), then make them equal.

1. Row 4 of Pascal's triangle: 1\;4\;6\;4\;1
2. In (a+2x)^4 the x^3 term is 4\times a\times(2x)^3=32ax^3
3. In (3-4x)^4 the x^2 term is 6\times3^2\times(-4x)^2
4. =6\times9\times16x^2=864x^2
5. 32a=864
6. Answer: a=27

Both coefficients contain a: square the whole term ax in the first bracket. Answer part (a) before attempting part (b).

Part (a)

1. In (2+ax)^4 the x^2 term is 6\times2^2\times(ax)^2=24a^2x^2
2. In (a+3x)^3 the x^2 term is 3\times a\times(3x)^2=27ax^2
3. 24a^2=27a
4. a\neq0, so divide both sides by a: 24a=27
5. a=\dfrac{27}{24}
6. Answer: a=\dfrac98

Part (b)

1. The coefficient is 27a
2. 27\times\dfrac98=\dfrac{243}{8}
3. (Check: 24a^2=24\times\dfrac{81}{64}=\dfrac{243}{8})
4. Answer: \dfrac{243}{8}

Square the whole term ax when you find the x^2 coefficient; for part (b) you need the x term of (2+ax)^4 as well as the x^2 term.

Part (a)

1. The x^2 term is 6\times2^2\times(ax)^2
2. (ax)^2=a^2x^2, so the coefficient is 24a^2
3. 24a^2=600
4. a^2=25, so a=5 or a=-5
5. a is negative, so a=-5

Part (b)

1. With a=-5: (2-5x)^4
2. Constant term: 2^4=16 (not needed)
3. x term: 4\times2^3\times(-5x)=-160x
4. x^2 term: 600x^2 (from part a)
5. In (1+3x)(\dots) the x^2 terms come from 1\times600x^2 and 3x\times(-160x)
6. 600-480=120
7. Answer: 120

Compare the constant terms first, then the terms in x. Answer part (a) before attempting part (b).

Part (a)

1. The constant term of the expansion is p^5
2. p^5=243
3. Answer: p=3

Part (b)

1. The x term is 5\times p^4\times qx
2. With p=3: 5\times81\times q=405q
3. 405q=-810
4. Answer: q=-2

Part (c)

1. The x^2 term is 10\times p^3\times(qx)^2
2. =10\times27\times(-2x)^2
3. =10\times27\times4x^2=1080x^2
4. Answer: r=1080

Write the coefficients of x and x^3 in terms of a, cubing the whole term ax for the second. Answer part (a) before attempting part (b).

Part (a)

1. Row 5 of Pascal's triangle: 1\;5\;10\;10\;5\;1
2. The x term is 5\times4^4\times ax=1280ax
3. The x^3 term is 10\times4^2\times(ax)^3=160a^3x^3
4. 160a^3=1280a
5. a\neq0, so divide by 160a: a^2=8
6. a=\pm\sqrt8
7. Answer: a=2\sqrt2 or a=-2\sqrt2

Part (b)

1. The x^2 term is 10\times4^3\times(ax)^2=640a^2x^2
2. From part (a), a^2=8
3. 640\times8=5120
4. Answer: 5120

When you add the two expansions the odd powers of x cancel. Answer part (a) before attempting part (b).

Part (a)

1. Row 4 of Pascal's triangle: 1\;4\;6\;4\;1
2. (x+2)^4=x^4+8x^3+24x^2+32x+16
3. (x-2)^4=x^4-8x^3+24x^2-32x+16
4. Add: the x^3 and x terms cancel
5. Answer: 2x^4+48x^2+32

Part (b)

1. 2x^4+48x^2+32=136
2. 2x^4+48x^2-104=0
3. Divide by 2: x^4+24x^2-52=0
4. This is a quadratic in x^2: (x^2+26)(x^2-2)=0
5. x^2=-26 has no solutions, so x^2=2
6. Answer: x=\sqrt2 or x=-\sqrt2