Applications: Polynomials and the Binomial Theorem

Each toss is a head or a tail, and all 2⁶ = 64 sequences of six tosses are equally likely. In the product of six brackets (1 + x)(1 + x) ⋯ (1 + x), choosing x from a bracket stands for a head and choosing 1 stands for a tail, so the binomial coefficients count the sequences.

1. Each of the 6 tosses has 2 outcomes, so there are 2⁶ = 64 sequences of heads and tails, and they are all equally likely.
2. By the binomial theorem, (1 + x)⁶ = 1 + 6x + 15x² + 20x³ + 15x⁴ + 6x⁵ + x⁶. The coefficients are row 6 of Pascal's triangle, 60, 61, …, 66.
3. Think of the six brackets as the six tosses: take x from a bracket for a head and 1 for a tail. A term in x² takes x from exactly two brackets, so its coefficient 62 = 6 × 52 × 1 = 15 is the number of sequences with exactly two heads.
4. (a) The probability of exactly two heads is 1564 ≈ 0.234.
5. At least five heads means five heads or six heads: 65 + 66 = 6 + 1 = 7 sequences. (b) The probability is 764 ≈ 0.109. Check: the whole row adds up to 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64, so every sequence is counted once.

answer(a) (1 + x)⁶ = 1 + 6x + 15x² + 20x³ + 15x⁴ + 6x⁵ + x⁶; 15 of the 64 sequences have exactly two heads, so the probability is 1564; (b) 764

techniqueThe Binomial Theorem

examsO-Level · GCSE Higher

Write q = 0.9 for the chance that a bulb works and p = 0.1 for the chance that it is faulty. The expansion of (q + p)⁴ has one term for each number of faulty bulbs, from 0 to 4, and its terms add up to 1⁴ = 1.

1. By the binomial theorem, (q + p)⁴ = q⁴ + 4q³p + 6q²p² + 4qp³ + p⁴, with q = 0.9 and p = 0.1. The whole sum is (0.9 + 0.1)⁴ = 1.
2. The term 4kq^4−kp^k is the chance of exactly k faulty bulbs: there are 4k ways to choose which k bulbs are faulty, and each of those ways has probability q^4−kp^k.
3. (a) The chance of exactly one faulty bulb is 4q³p = 4 × 0.729 × 0.1 = 0.2916.
4. The chance of no faulty bulb is q⁴ = 0.9⁴ = 0.6561, so the chance of at most one faulty bulb is 0.6561 + 0.2916 = 0.9477.
5. (b) A box is returned with probability 1 − 0.9477 = 0.0523. Check by adding the last three terms: 6 × 0.81 × 0.01 + 4 × 0.9 × 0.001 + 0.0001 = 0.0486 + 0.0036 + 0.0001 = 0.0523.

answer(a) (0.9 + 0.1)⁴ = 0.6561 + 0.2916 + 0.0486 + 0.0036 + 0.0001, and the probability of exactly one faulty bulb is 0.2916; (b) 0.0523

techniqueThe Binomial Theorem

examsO-Level · GCSE Higher

The binomial theorem expands (1 + x)¹⁰ into terms in rising powers of x. When x is as small as 0.02, each power is fifty times smaller than the one before, so the first few terms give the balance very closely.

1. By the binomial theorem, (1 + x)¹⁰ = 1 + 10x + 102x² + 103x³ + ⋯ = 1 + 10x + 45x² + 120x³ + ⋯.
2. Put x = 0.02: 10x = 0.2, 45x² = 45 × 0.0004 = 0.018 and 120x³ = 120 × 0.000008 = 0.00096.
3. The first four terms add up to 1 + 0.2 + 0.018 + 0.00096 = 1.21896. (a) The balance is about 1000 × 1.21896 = $1218.96.
4. The calculator gives 1000 × 1.218994 = $1218.99 to the nearest cent, so the estimate is $0.03 short: 3 cents.
5. (b) The next term is 104x⁴ = 210 × 0.02⁴ = 210 × 0.00000016 = 0.0000336. On $1000 it is worth 1000 × 0.0000336 = $0.0336, which is 3.36 cents, almost the whole error. The terms after it are smaller still.

answer(a) 1 + 0.2 + 0.018 + 0.00096 = 1.21896, so about $1218.96; (b) the balance is $1218.99, so the estimate is 3 cents short, and the next term, 210 × 0.02⁴ = 0.0000336, is worth 3.36 cents

techniqueThe Binomial Theorem

examsO-Level · GCSE Higher

When every length is multiplied by 1 + x, the volume, a product of three lengths, is multiplied by (1 + x)³. For a small x the terms in x² and x³ are tiny, so a small percentage growth in every length makes about three times that percentage growth in the volume.

1. Every length is multiplied by 1 + x, where x = 0.02. The volume is multiplied by (1 + x)³ = 1 + 3x + 3x² + x³.
2. For a small x, the terms 3x² and x³ are tiny, so (1 + x)³ ≈ 1 + 3x = 1 + 0.06 = 1.06. The volume grows by about 3 × 2% = 6%.
3. (a) The volume was 10 × 8 × 5 = 400 cm³, so the new volume is about 400 × 1.06 = 424 cm³, an increase of about 6%.
4. Exactly, 1.02³ = 1 + 0.06 + 0.0012 + 0.000008 = 1.061208, so the new volume is 400 × 1.061208 = 424.4832, which is 424.48 cm³. Check with the new lengths: 10.2 × 8.16 × 5.1 = 424.4832.
5. (b) The estimate is short by 424.48 − 424 = 0.48 cm³. The term 3x² is worth 400 × 0.0012 = 0.48 cm³ of it, and the term x³ only 400 × 0.000008 = 0.0032 cm³.

answer(a) (1 + x)³ = 1 + 3x + 3x² + x³; an increase of about 6%, to about 424 cm³; (b) 424.48 cm³, so the estimate is 0.48 cm³ short, almost all of it the term 3x²

techniqueThe Binomial Theorem

examsO-Level · GCSE Higher

For an index n that is not a positive whole number, the binomial series (1 + x)ⁿ = 1 + nx + n(n − 1)2!x² + ⋯ never stops. Its sum settles on the true value only when −1 < x < 1.

1. The binomial series for any index n is (1 + x)ⁿ = 1 + nx + n(n − 1)2!x² + ⋯. With n = 12, the coefficient of x is 12 and the coefficient of x² is 12 × (−12)2 = −18.
2. So (1 + x)¹² = 1 + 12x − 18x² + ⋯. The index is not a whole number, so the series never stops, and it is valid only for −1 < x < 1.
3. (a) The length grows by 4%, so x = 0.04: √1.04 ≈ 1 + 0.02 − 0.00168 = 1 + 0.02 − 0.0002 = 1.0198. A calculator gives 1.019804, so the estimate is right to 4 decimal places.
4. (b) One swing now takes about 2 × 1.0198 = 2.0396 seconds, about 0.04 seconds longer than before, so the clock runs slow.
5. For √3 = (1 + 2)¹² the value x = 2 lies outside −1 < x < 1. The terms of the series are then 1, 1, −0.5, 0.5, −0.625, 0.875, …, which grow instead of shrinking, so their sum never settles and the series cannot be used.

answer(a) (1 + x)¹² = 1 + 12x − 18x² + ⋯, valid for −1 < x < 1, so √1.04 ≈ 1.0198; (b) about 2.0396 seconds; x = 2 lies outside −1 < x < 1, where the terms grow and the series does not converge

techniqueBinomial Series for a Negative or Fractional Index

examsH2

The binomial series expands (1 + u)ⁿ, with a 1 in the bracket. So the 50 is taken out of (50 + v)⁻¹ first, which leaves (1 + v50)⁻¹. The series holds for −1 < v50 < 1, so the range for v is fifty times as wide.

1. Take the 50 out of the bracket first: T = 100(50 + v)⁻¹ = 100 × 50⁻¹(1 + v50)⁻¹ = 2(1 + v50)⁻¹.
2. The binomial series with n = −1 is (1 + u)⁻¹ = 1 − u + u² − ⋯, valid for −1 < u < 1. Put u = v50: T = 2(1 − v50 + v²2500 − ⋯).
3. (a) T ≈ 2 − v25 + v²1250. It is valid for −1 < v50 < 1, which is −50 < v < 50: any current slower than the ferry itself.
4. With v = 5: T ≈ 2 − 525 + 251250 = 2 − 0.2 + 0.02 = 1.8200 hours.
5. (b) The exact time is 10055 = 1.8182 hours to 4 decimal places. The estimate is 0.0018 hours too long, about 6.5 seconds. The next term of the series, 2 × (−(550)³) = −0.002, accounts for it.

answer(a) T ≈ 2 − v25 + v²1250, valid for −50 < v < 50; (b) about 1.8200 hours, against an exact 1.8182 hours, so the estimate is about 6.5 seconds too long

techniqueExpanding (a + bx) to a Rational Power · Binomial Series for a Negative or Fractional Index

examsH2

The remainder when a polynomial is divided by (x − k) is its value at x = k. When that value is 0, (x − k) is a factor. Each fact about the trail is one equation in a and b.

1. The trail is level with the car park at x = 1, so y = 0 there, and by the factor theorem (x − 1) is a factor: 1 + a + b + 6 = 0, which gives a + b = −7.
2. By the remainder theorem, the remainder when the cubic is divided by (x − 2) is its value at x = 2, which is −4 because the trail is below the car park: 8 + 4a + 2b + 6 = −4, so 4a + 2b = −18 and 2a + b = −9.
3. Subtract the first equation from the second: a = −2, and then b = −7 − (−2) = −5. (a) a = −2 and b = −5, so y = x³ − 2x² − 5x + 6.
4. Divide by the factor (x − 1): x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) = (x − 1)(x − 3)(x + 2).
5. (b) The trail is level with the car park again at x = 3, which is 30 m along. The root x = −2 lies before the start of the trail, outside 0 ≤ x ≤ 3.5, so it is rejected. Check: at x = 3, 27 − 18 − 15 + 6 = 0.

answer(a) a = −2 and b = −5, so y = x³ − 2x² − 5x + 6; (b) at x = 3, 30 m along the trail; the root x = −2 lies outside 0 ≤ x ≤ 3.5

techniqueFactor and Remainder Theorem · Solving a Cubic Completely

examsO-Level · SAT

The rod fixes the height in terms of x, and the volume then gives a cubic in x. One root found by the factor theorem splits off a linear factor; dividing leaves a quadratic, which gives the other two roots. Each root is then checked against the frame.

1. The base and the top are squares with 4 edges of x m each, and the four upright edges are h m each, so 8x + 4h = 28 and h = 7 − 2x. The height must be more than 0, so 0 < x < 3.5.
2. The volume is x²(7 − 2x) = 9, so 7x² − 2x³ = 9, which rearranges to 2x³ − 7x² + 9 = 0.
3. (a) Try the factors of 9 over the factors of 2. At x = 1 the cubic is 2 − 7 + 9 = 4, not 0. At x = −1 it is −2 − 7 + 9 = 0, so by the factor theorem (x + 1) is a factor and x = −1 is a root.
4. Divide by (x + 1): 2x³ − 7x² + 9 = (x + 1)(2x² − 9x + 9), and the quadratic factorizes as (2x − 3)(x − 3). The roots are x = −1, x = 1.5 and x = 3.
5. (b) Reject x = −1, because a length cannot be negative. x = 3 gives h = 7 − 6 = 1, a frame 3 m by 3 m by 1 m, and x = 1.5 gives h = 7 − 3 = 4, a frame 1.5 m by 1.5 m by 4 m. Check: 3² × 1 = 9 and 1.5² × 4 = 9.

answer(a) x = −1 is a root, so (x + 1) is a factor; (b) x = −1, 1.5 or 3; rejecting −1, the frames are 3 m by 3 m by 1 m and 1.5 m by 1.5 m by 4 m

techniqueSolving a Cubic Completely · Factor and Remainder Theorem

examsO-Level

A cubic x³ + px² + qx + r = 0 with roots a, b and c has a + b + c = −p, ab + bc + ca = q and abc = −r. The edges, the card and the volume give exactly these three sums, so the cubic can be written down, and the diagonal found from the sums alone.

1. Call the dimensions a, b and c cm. The box has 4 edges of each length, so 4(a + b + c) = 48 and a + b + c = 12. It has two faces of each size, so 2(ab + bc + ca) = 94 and ab + bc + ca = 47. The volume gives abc = 60.
2. A cubic with roots a, b and c is x³ − (a + b + c)x² + (ab + bc + ca)x − abc = 0, which is x³ − 12x² + 47x − 60 = 0.
3. The rod from corner to opposite corner has length √a² + b² + c², and a² + b² + c² = (a + b + c)² − 2(ab + bc + ca) = 144 − 94 = 50. (a) The longest rod is √50 = 5√2 ≈ 7.07 cm.
4. Try the factors of 60: at x = 3 the cubic is 27 − 108 + 141 − 60 = 0, so (x − 3) is a factor. Dividing, x³ − 12x² + 47x − 60 = (x − 3)(x² − 9x + 20) = (x − 3)(x − 4)(x − 5).
5. (b) The box is 3 cm by 4 cm by 5 cm. Check: 3 + 4 + 5 = 12, 12 + 20 + 15 = 47, 3 × 4 × 5 = 60, and √9 + 16 + 25 = √50.

answer(a) the longest rod is √50 = 5√2 ≈ 7.07 cm; (b) 3 cm by 4 cm by 5 cm

techniqueRoots and Coefficients of a Cubic · Solving a Cubic Completely

examsO-Level

If every new root is z = x + 2, then x = z − 2, and putting z − 2 in place of x gives an equation in z with every root moved 2 later. Halving works the same way, with x = 2y. The coefficients of the new equation give the sum and the product of the new roots.

1. A time z on the second clock is 2 seconds more than the time x: z = x + 2, so x = z − 2. Substitute into the cubic: (z − 2)³ − 9(z − 2)² + 23(z − 2) − 15 = 0.
2. Expand each bracket: z³ − 6z² + 12z − 8 − 9z² + 36z − 36 + 23z − 46 − 15 = 0. (a) Collect the terms: z³ − 15z² + 71z − 105 = 0.
3. On the faster setting a new time is y = x2, so x = 2y. Substitute: 8y³ − 36y² + 46y − 15 = 0, which already has whole-number coefficients.
4. (b) The equation is 8y³ − 36y² + 46y − 15 = 0, and the sum of its roots is 368 = 4.5 seconds, half of the 9 seconds from the first cubic.
5. Check: x = 1 gives 1 − 9 + 23 − 15 = 0, and dividing out (x − 1) leaves x² − 8x + 15 = (x − 3)(x − 5). The cart passes at 1, 3 and 5 seconds; on the second clock at 3, 5 and 7, which add up to 15 and multiply to 105; on the faster setting at 0.5, 1.5 and 2.5, which add up to 4.5.

answer(a) z³ − 15z² + 71z − 105 = 0; (b) 8y³ − 36y² + 46y − 15 = 0, whose roots add up to 4.5 seconds; the times are 1, 3 and 5 seconds, then 3, 5 and 7 seconds, then 0.5, 1.5 and 2.5 seconds

techniqueTransforming the Roots of a Polynomial · Roots and Coefficients of a Cubic