Edexcel International GCSE Maths Solutions: 4WM1H/01R May 2025 (Unit 1 Higher, R paper)
Full worked solutions to the Pearson Edexcel International GCSE Mathematics A (Modular) Unit 1 Higher Tier R paper from May 2025 (4WM1H/01R, sat Thursday 15 May 2025): 24 questions, 100 marks, 2 hours, calculator allowed. The R paper is the alternative version of 4WM1H/01 that Pearson sets for some regions. Attempt each question yourself first, then click “Show solution” to see the working, with a short explanation of why each step is done.
The length of a ship is 142.8 m, correct to 1 decimal place.
(i) Write down the lower bound of the length of the ship.
(ii) Write down the upper bound of the length of the ship.
Show solution
- (i) Correct to 1 decimal place means ± 0.05: 142.8 − 0.05 = 142.75 mA value rounded to 1 d.p. can be up to half of 0.1 (0.05) away from 142.8. The lower bound is the smallest value that would round to 142.8.
- (ii) 142.8 + 0.05 = 142.85 mThe upper bound is half a unit above. Strictly, 142.85 itself would round up to 142.9, but by convention the upper bound is written as 142.85.
Show that 214 × 157 = 367
Show solution
- Step 1: convert to improper fractions. 214 = 94 and 157 = 127You can’t multiply mixed numbers directly. Multiply the whole number by the denominator and add the numerator: 2 × 4 + 1 = 9 and 1 × 7 + 5 = 12.
- Step 2: multiply. 94 × 127 = 10828 = 277 (or cancel 4 into 12 first: 9 × 37 = 277)Multiply the tops together and the bottoms together. Cancelling the common factor of 4 first keeps the numbers small.
- Step 3: convert back. 27 ÷ 7 = 3 remainder 6, so 277 = 367 ✓The question gives the answer as a mixed number, so convert back to match it. 3 × 7 = 21, leaving 6 sevenths.
Here is a biased 5-sided spinner. When the spinner is spun, it can land on blue or on green or on purple or on red or on yellow. The table gives information about the probability of the spinner landing on each colour.
| Colour | blue | green | purple | red | yellow |
|---|---|---|---|---|---|
| Probability | 0.12 | 0.20 | 0.38 | 4x | x |
Sophie spins the spinner once.
(a) Work out the probability that the spinner lands on blue or on green or on purple.
Max spins the spinner 350 times.
(b) Work out an estimate for the number of times the spinner lands on red.
Show solution
- (a) 0.12 + 0.20 + 0.38 = 0.70The spinner can only land on one colour at a time, so for “blue or green or purple” add the probabilities.
- (b) Step 1: All five probabilities add to 1, so 4x + x = 1 − 0.70 = 0.30The spinner must land on one of the five colours, so the total probability is 1. Part (a) already gives the sum of the first three.
- Step 2: 5x = 0.30 → x = 0.06Collect like terms (4x + x = 5x), then divide by 5.
- Step 3: P(red) = 4x = 4 × 0.06 = 0.24The table gives red as 4x, so substitute x = 0.06.
- Step 4: Estimate = 0.24 × 350 = 84Expected frequency = probability × number of trials.
ℰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A = {2, 4, 6, 8, 10, 12} B = {3, 6, 9, 12} C = {1, 3, 5, 7, 9, 11}
(a) List the members of the set (i) A ∪ B (ii) B′
(b) Write a symbol from the box on each dotted line to make each of the following a true statement.
| ℰ | ∩ | ∪ | Ø | ∈ | ∉ |
(i) A ∩ C = …….. (ii) 13 …….. ℰ
Show solution
- (a)(i) 2, 3, 4, 6, 8, 9, 10, 12∪ (union) means in A or B or both. List every number that appears in either set, writing 6 and 12 only once.
- (a)(ii) 1, 2, 4, 5, 7, 8, 10, 11B′ (the complement) means everything in ℰ that is not in B. Cross 3, 6, 9 and 12 off the list 1 to 12.
- (b)(i) A ∩ C = ØA is the even numbers and C is the odd numbers, so no number is in both. An intersection with nothing in it is the empty set, Ø.
- (b)(ii) 13 ∉ ℰ∈ means “is a member of” and ∉ means “is not a member of”. The universal set only goes up to 12, so 13 is not in it.
38 × 34 = 3m (a) Find the value of m
412 ÷ 47 = 4n (b) Find the value of n
(62)10 = 6p (c) Find the value of p
5r = 1 (d) Write down the value of r
Show solution
- (a) 8 + 4, so m = 12When multiplying powers of the same base, add the indices.
- (b) 12 − 7, so n = 5When dividing powers of the same base, subtract the indices.
- (c) 2 × 10, so p = 20For a power of a power, multiply the indices: (6²)¹⁰ is ten lots of 6² multiplied together.
- (d) r = 0Any non-zero number to the power 0 equals 1. By the index laws, 5ⁿ ÷ 5ⁿ = 5⁰, and anything divided by itself is 1.
The diagram shows triangle ABC.
CA = (4x + 6) cm, CB = (30 − x) cm, angle CAB = angle CBA. Work out the value of x.
Show solution
- Step 1: The base angles at A and B are equal, so the triangle is isosceles and CA = CB.In an isosceles triangle, the two equal sides are opposite the two equal angles. Angle A is opposite CB and angle B is opposite CA.
- Step 2: 4x + 6 = 30 − xWriting the two equal sides equal to each other gives an equation in x.
- Step 3: 5x = 24 → x = 4.8Add x to both sides and subtract 6 from both sides, then divide by 5. Check: 4(4.8) + 6 = 25.2 and 30 − 4.8 = 25.2 ✓
Change a speed of 40 metres per second to a speed in kilometres per hour. Show your working clearly.
Show solution
- Step 1: metres per hour. 40 × 3600 = 144 000 m per hourThere are 60 × 60 = 3600 seconds in an hour. If you travel 40 m every second, in an hour you travel 3600 times as far.
- Step 2: kilometres per hour. 144 000 ÷ 1000 = 144 km/h1 km = 1000 m, so divide the number of metres by 1000.
ABD and BCD are right-angled triangles (right angles at A and B). Angle ABD = 65°, angle CDB = 32°, BC = 17 cm. Work out the length of AD. Give your answer correct to 3 significant figures.
Show solution
- Step 1: find BD in triangle BCD. tan 32° = 17BD → BD = 17tan 32° = 27.205… cmAD is in triangle ABD, but we only know an angle there. BD is the side the two triangles share, so find it first. In triangle BCD (right angle at B), BC = 17 is opposite the 32° angle and BD is adjacent, so use tan (TOA).
- Step 2: find AD in triangle ABD. sin 65° = ADBD → AD = 27.205… × sin 65° = 24.656…In triangle ABD (right angle at A), BD is the hypotenuse and AD is opposite the 65° angle at B, so use sin (SOH): opposite = hypotenuse × sin.
- AD = 24.7 cm (3 s.f.)Keep the unrounded BD in your calculator, and round only at the end.
(a) Simplify (5k3n4)2
(b) Factorise fully 12m2y4 + 18m3y
(c) (i) Factorise x2 − 13x + 42 (ii) Hence, solve x2 − 13x + 42 = 0
Show solution
- (a) 5² = 25, (k³)² = k6, (n4)² = n8
Answer: 25k⁶n⁸The power outside the bracket applies to every factor inside, including the 5. For a power of a power, multiply the indices. - (b) The highest common factor is 6m²y:
12m²y4 ÷ 6m²y = 2y³ and 18m³y ÷ 6m²y = 3m
Answer: 6m²y(2y³ + 3m)“Fully” means take out the biggest common factor: the HCF of 12 and 18 (6), and the lowest power of each letter in both terms (m² and y). Divide each term by it to find what goes in the bracket. - (c)(i) Two numbers that multiply to +42 and add to −13: −6 and −7
Answer: (x − 6)(x − 7)To factorise x² + bx + c, find two numbers that multiply to c and add to b. The product is positive and the sum negative, so both numbers are negative. - (c)(ii) x = 6 or x = 7“Hence” means use part (i). If two brackets multiply to 0, one of them must be 0, so x − 6 = 0 or x − 7 = 0.
The diagram shows a rectangular sheet of metal.
BC = 8 cm, DB = 17 cm. The sheet of metal lies flat on a table. The force exerted by the sheet of metal on the table is 150 newtons. Work out the pressure on the table due to the sheet of metal. Show your working clearly.
pressure = forcearea
Show solution
- Step 1: find DC using Pythagoras. DC = √(17² − 8²) = √(289 − 64) = √225 = 15 cmTo find the area we need both sides of the rectangle. The diagonal DB makes a right-angled triangle BCD with hypotenuse 17, so subtract the squares to find the missing side.
- Step 2: area. 15 × 8 = 120 cm²The whole sheet touches the table, so the contact area is the area of the rectangle: length × width.
- Step 3: pressure. 150120 = 1.25 N/cm²Substitute into pressure = force ÷ area. The area is in cm², so the pressure comes out in newtons/cm², which matches the answer line.
The probability that it will snow on Saturday is 0.2. If it snows on Saturday, the probability that it will snow on Sunday is 0.35. If it does not snow on Saturday, the probability that it will snow on Sunday is 0.4.
(a) Use this information to complete the probability tree diagram.
(b) Work out the probability that it will snow on both Saturday and Sunday.
Show solution
- (a) Saturday: snow 0.2, not snow 0.8
After snow: snow 0.35, not snow 0.65
After no snow: snow 0.4, not snow 0.6Each pair of branches from one point must add to 1, because it either snows or it doesn’t. Here Sunday’s probabilities depend on Saturday, so the two pairs of Sunday branches are different. - (b) 0.2 × 0.35 = 0.07“Snow on Saturday AND snow on Sunday” means follow the top path and multiply along its branches. Use 0.35, not 0.4, because it has already snowed on Saturday.
(a) Solve 5a + 83 − 2a + 54 = 23 Show clear algebraic working.
(b) Express (√y3)−1 in the form cyn where c and n are numbers to be found.
Show solution
- (a) Step 1: multiply every term by 12. 4(5a + 8) − 3(2a + 5) = 27612 is the lowest common multiple of 3 and 4, so multiplying by 12 clears both fractions. 12 ÷ 3 = 4 and 12 ÷ 4 = 3. Don’t forget to multiply the 23 on the right by 12 as well.
- Step 2: expand. 20a + 32 − 6a − 15 = 276The minus sign in front of the second bracket multiplies both terms: −3 × 5 = −15, not +15.
- Step 3: simplify and solve. 14a + 17 = 276 → 14a = 259 → a = 18.5Collect like terms (20a − 6a and 32 − 15), subtract 17 from both sides, then divide by 14.
- (b) Step 1: flip the fraction. (√y3)−1 = 3√yA power of −1 means the reciprocal, so the fraction turns upside down.
- Step 2: write the root as a power. √y = y½, so 3y½ = 3y^(−½) (c = 3, n = −½)A square root is the power ½. Dividing by y^½ is the same as multiplying by y^(−½), which gives the required form cyⁿ.
In the diagram, AB represents a vertical tower and CD represents a vertical building.
The bottom of the tower, B, and the bottom of the building, D, are on horizontal ground such that BD = 25 m. The angle of elevation of A from C is 30°. The angle of depression of B from C is 42°.
(a) Work out the height of the tower AB. Give your answer correct to one decimal place.
(b) Work out the angle of elevation of A from D. Give your answer correct to one decimal place.
Show solution
- (a) Step 1: the part of the tower below C‘s level. 25 × tan 42° = 22.510… mThe horizontal line from C to the tower is 25 m long (the same as BD). With the 42° angle of depression, the drop down to B is opposite the angle and the 25 m is adjacent, so use tan. This drop also equals the building’s height CD.
- Step 2: the part of the tower above C‘s level. 25 × tan 30° = 14.433… mThe same idea upwards: the rise to A is opposite the 30° angle of elevation, with the same 25 m adjacent side.
- Step 3: add. AB = 22.510… + 14.433… = 36.943… = 36.9 mThe horizontal line through C splits the tower into these two parts, so the full height is their sum.
- (b) tan θ = 36.943…25 → θ = tan⁻¹(1.4777…) = 55.913… = 55.9°From D on the ground, triangle ABD is right-angled at B. AB is opposite the angle at D and BD = 25 is adjacent, so use tan, then inverse tan to get the angle. Use the unrounded AB.
Use algebra to show that the recurring decimal 0.612 = 101165 (the digits 1 and 2 recur: 0.6121212…)
Show solution
- Step 1: Let x = 0.6121212…Giving the decimal a name lets us do algebra with it.
- Step 2: 1000x = 612.121212… and 10x = 6.121212…We need two multiples of x with exactly the same recurring part after the decimal point. Multiplying by 10 moves the 6 past the point; multiplying by 1000 moves one more full “12” cycle past it.
- Step 3: subtract. 1000x − 10x = 612.1212… − 6.1212… → 990x = 606The recurring parts are identical, so they cancel completely, leaving whole numbers.
- Step 4: x = 606990 = 101/165 ✓Divide by 990, then simplify by dividing the top and bottom by 6 (606 ÷ 6 = 101 and 990 ÷ 6 = 165).
(a) Solve 3x2 + 4x − 18 = 0. Show your working clearly. Give your solutions correct to 3 significant figures.
(b) Expand and simplify (2x − 3)(x + 4)(x − 1)
Show solution
- (a) Step 1: Quadratic formula with a = 3, b = 4, c = −18:
x = −4 ± √(4² − 4 × 3 × (−18))2 × 3 = −4 ± √2326“3 significant figures” tells you the quadratic won’t factorise, so use the formula from the formulae sheet. −4 × 3 × (−18) = +216, so the number under the root is 16 + 216 = 232. - Step 2: √232 = 15.231…
x = −4 + 15.231…6 = 1.8719… or x = −4 − 15.231…6 = −3.2052…The ± gives two solutions. Work out each one separately. - x = 1.87 or x = −3.21Round each to 3 significant figures. −3.2052… rounds to −3.21 because the fourth figure is 5.
- (b) Step 1: (x + 4)(x − 1) = x² − x + 4x − 4 = x² + 3x − 4Expand two brackets at a time. Multiply every term in the first bracket by every term in the second, then collect like terms.
- Step 2: (2x − 3)(x² + 3x − 4) = 2x³ + 6x² − 8x − 3x² − 9x + 12Multiply 2x by each of the three terms, then −3 by each of the three terms. That gives six terms in total.
- Step 3: collect like terms. 2x³ + 3x² − 17x + 126x² − 3x² = 3x² and −8x − 9x = −17x.
√96 = 2√k (a) Find the value of k
(b) Show that 83 − √5 can be written in the form a + √b where a and b are integers. Show each stage of your working.
Show solution
- (a) 2√k = √(4k), so 4k = 96 and k = 24Move the 2 inside the root by squaring it (2 = √4). Then the two square roots are equal, so the numbers inside are equal. Check: √96 = √(4 × 24) = 2√24 ✓
- (b) Step 1: rationalise. 83 − √5 × 3 + √53 + √5 = 8(3 + √5)(3 − √5)(3 + √5)To remove the surd from the denominator, multiply the top and bottom by its conjugate (the same terms with the opposite sign). This is multiplying by 1, so the value doesn’t change.
- Step 2: denominator. (3 − √5)(3 + √5) = 9 − 5 = 4(a − b)(a + b) = a² − b², and (√5)² = 5, so the surds cancel.
- Step 3: simplify. 8(3 + √5)4 = 2(3 + √5) = 6 + 2√58 ÷ 4 = 2, then multiply out the bracket.
- Step 4: single surd. 2√5 = √(4 × 5) = √20, so the answer is 6 + √20 (a = 6, b = 20) ✓The form a + √b needs one surd with no number in front, so move the 2 inside the root by squaring it.
Kannika has 9 counters. There is a number on each counter: 1, 1, 2, 3, 3, 3, 5, 6, 6. Kannika puts the 9 counters in a bag. She takes at random a counter from the bag and does not replace the counter. She then takes at random a second counter from the bag. Work out the probability that the sum of the numbers on the two counters is less than 5.
Show solution
- Step 1: list the pairs that sum to less than 5. 1 + 1 = 2, 1 + 2 = 3 and 1 + 3 = 4. (2 + 2 is impossible, because there is only one 2.)Every pair with a 5 or a 6 is too big, and 2 + 3 = 5 is not less than 5. So every successful pair must contain a 1.
- Step 2: probability of each outcome (without replacement, so the second denominator is 8):For each order, multiply the probability of the first counter by the probability of the second (“first AND then second”). 1 + 2 and 1 + 3 can happen in either order, so both orders are counted.
- 1 then 1: 29 × 18 = 272
- 1 then 2, or 2 then 1: 29 × 18 + 19 × 28 = 472
- 1 then 3, or 3 then 1: 29 × 38 + 39 × 28 = 1272
- Step 3: add. 2 + 4 + 1272 = 1872 = 1/4These outcomes can’t happen together, so add their probabilities (‘or’ means add). Simplify by dividing the top and bottom by 18.
(a) Express 2x2 − 12x + 23 in the form a(x − b)2 + c where a, b and c are integers. Show clear algebraic working.
(b) Hence, state the least value of 2x2 − 12x + 23
Show solution
- (a) Step 1: factor 2 out of the x terms. 2(x² − 6x) + 23Completing the square is easiest when the x² coefficient is 1. Only take the 2 out of the x terms, and leave the 23 outside.
- Step 2: complete the square. x² − 6x = (x − 3)² − 9Halve the coefficient of x (−6 ÷ 2 = −3) to make the bracket. (x − 3)² = x² − 6x + 9 has an extra +9, so subtract it.
- Step 3: substitute back. 2[(x − 3)² − 9] + 23 = 2(x − 3)² − 18 + 23 = 2(x − 3)² + 5Multiply the 2 into the square bracket: 2 × (−9) = −18, then −18 + 23 = 5. So a = 2, b = 3 and c = 5.
- (b) Least value = 5 (when x = 3)A square can never be negative, so 2(x − 3)² is at least 0. It equals 0 when x = 3, which leaves just the + 5.
The diagram shows a pentagon ABCDE.
ABCDE is made from a square ABDE and a triangle BCD. M is the midpoint of BD. AB = (2x + 3) cm, MC = 2x cm, angle BMC = 90°. The area of pentagon ABCDE is 84 cm². Work out the area of square ABDE. Show clear algebraic working.
Show solution
- Step 1: area of the square. (2x + 3)² = 4x² + 12x + 9The square has side 2x + 3, and its area is side × side. Expand the bracket by multiplying it by itself.
- Step 2: area of the triangle. ½ × (2x + 3) × 2x = 2x² + 3xThe triangle’s base is BD, which is a side of the square (2x + 3). MC is perpendicular to BD, so it is the perpendicular height.
- Step 3: form an equation. 4x² + 12x + 9 + 2x² + 3x = 84 → 6x² + 15x − 75 = 0 → 2x² + 5x − 25 = 0The pentagon is the square plus the triangle, so their areas add to 84. Collect like terms, subtract 84, then divide everything by 3 to make the numbers smaller.
- Step 4: solve. (2x − 5)(x + 5) = 0 → x = 2.5 or x = −5Factorise: we need two numbers that multiply to 2 × (−25) = −50 and add to 5, which are 10 and −5. x = −5 is rejected because it would make the length MC = 2x negative.
- Step 5: area of the square. Side = 2(2.5) + 3 = 8 cm, so area = 8² = 64 cm²The question asks for the area of the square, not x. Check: the triangle is ½ × 8 × 5 = 20, and 64 + 20 = 84 ✓
The diagram shows cuboid ABCDEFGH.
AB = 9 cm, AF = 7 cm, FC = 18 cm. Calculate the length of BC. Give your answer correct to 3 significant figures.
Show solution
- Step 1: use the 3D Pythagoras rule. FC is the diagonal through the cuboid, so FC² = AB² + BC² + AF²F is above A and C is the opposite bottom corner. The diagonal FC is the hypotenuse of triangle FAC (right angle at A), and AC is the hypotenuse of triangle ABC on the base. Combining the two gives FC² = AB² + BC² + AF²: the cuboid’s three dimensions, squared and added.
- Step 2: substitute. 18² = 9² + BC² + 7² → 324 = 81 + BC² + 49Put in the three known lengths, leaving BC as the unknown.
- Step 3: solve. BC² = 324 − 81 − 49 = 194 → BC = √194 = 13.928… = 13.9 cmSubtract the known squares from 324, then take the square root. Round to 3 significant figures.
The histogram shows information about the ages of the members of a chess club.
There are 60 members aged between 10 and 25. There are no members older than 70. Work out an estimate for the number of members who are over 35 years of age.
Show solution
- Step 1: find the frequency density scale. The 10–25 bar has 60 members and width 15, so its frequency density = 60 ÷ 15 = 4. That bar is 40 small squares tall, so 1 small square = 0.1.The vertical axis has no numbers on it, so use the one known frequency to work out the scale. Frequency density = frequency ÷ class width.
- Step 2: read the bars you need. 30–40 bar: 10 small squares = FD 1.0. 40–70 bar: 7 small squares = FD 0.7.Over 35 means the right half of the 30–40 bar and the whole of the 40–70 bar, so only these two heights are needed.
- Step 3: frequency = frequency density × width. 35–40: 5 × 1.0 = 5. 40–70: 30 × 0.7 = 21.In a histogram, the area of a bar represents frequency. 35 falls in the middle of the 30–40 bar, so use only its 5-year width from 35 to 40, assuming the members are spread evenly.
- Step 4: add. 5 + 21 = 26The question says “estimate” because of the even-spread assumption inside the 30–40 bar.
BAC is a sector of a circle, centre A. BCD is a sector of a circle, centre C. Angle BAC = 40°. Angle BCD = 130°. Area of shaded segment = 28 cm². Find the length of the arc BD. Give your answer correct to 3 significant figures.
Show solution
- Step 1: segment area formula. Let r = AB = AC. Segment = sector − triangle = 40360πr² − ½r² sin 40°The shaded segment lies between chord BC and the arc of the circle centred A. It is the sector BAC with triangle BAC removed. The triangle has two sides r with a 40° angle between them, so its area is ½ab sin C.
- Step 2: solve for r. r²(0.34906… − 0.32139…) = 28 → r² × 0.027670… = 28 → r² = 1011.9… → r = 31.809… cmBoth terms contain r², so factorise it out. 40/360 × π = 0.34906… and ½ sin 40° = 0.32139…. The difference is small, so keep plenty of decimal places. Divide 28 by their difference, then square-root.
- Step 3: find BC. Cosine rule in triangle BAC: BC² = r² + r² − 2r² cos 40° → BC = 21.759… cmBC is the radius of the second sector (centre C), so we need its length. We know two sides of triangle BAC and the angle between them, so use the cosine rule. (Equivalently, BC = 2r sin 20°, because the isosceles triangle splits into two right-angled triangles.)
- Step 4: arc length. Arc BD = 130360 × 2π × 21.759… = 49.369…An arc is a fraction θ/360 of the full circumference 2πr. Here the radius is CB and the angle at C is 130°.
- Arc BD = 49.4 cm (3 s.f.)Keep every value unrounded in your calculator until this last step, because errors build up over four stages.
A circle C has centre (3q, 4q) where q is an integer. The straight line with equation y = −54x + 214 is the tangent to C at the point (p, p + 3) where p is an integer. Work out the radius of C. Give your answer in the form √a where a is an integer. Show your working clearly.
Show solution
- Step 1: find the point of contact. (p, p + 3) lies on the tangent, so p + 3 = −54p + 214 → 4p + 12 = −5p + 21 → 9p = 9 → p = 1. The point is (1, 4).A point on a line satisfies its equation. Substitute x = p and y = p + 3, then multiply by 4 to clear the fractions.
- Step 2: gradient of the radius. The tangent’s gradient is −54, so the radius to (1, 4) has gradient 45.A tangent is perpendicular to the radius at the point of contact. Perpendicular gradients multiply to −1, so take the negative reciprocal of −5/4.
- Step 3: find q. The radius joins (1, 4) to the centre (3q, 4q): 4q − 43q − 1 = 45 → 20q − 20 = 12q − 4 → 8q = 16 → q = 2. The centre is (6, 8).Gradient = change in y ÷ change in x between the two ends of the radius. Set it equal to 4/5 and cross-multiply.
- Step 4: radius. Distance from (6, 8) to (1, 4) = √((6 − 1)² + (8 − 4)²) = √(25 + 16) = √41The radius is the distance from the centre to the point of contact. Use Pythagoras on the horizontal and vertical differences.
Write (2x − 1) + 2x2 − 5x − 12x2 − 16 × x2 + 4x6x2 + 11x + 3 as a single fraction in its simplest form. Show clear algebraic working.
Show solution
- Step 1: factorise every part.
- 2x² − 5x − 12 = (2x + 3)(x − 4)
- x² − 16 = (x − 4)(x + 4)
- x² + 4x = x(x + 4)
- 6x² + 11x + 3 = (2x + 3)(3x + 1)
- Step 2: multiply and cancel. (2x + 3)(x − 4)(x − 4)(x + 4) × x(x + 4)(2x + 3)(3x + 1) = x3x + 1(2x + 3), (x − 4) and (x + 4) each appear on both the top and the bottom, so they cancel. Do the multiplication before the addition, following the order of operations.
- Step 3: add (2x − 1). (2x − 1)(3x + 1)3x + 1 + x3x + 1 = 6x² − x − 1 + x3x + 1To add, write 2x − 1 as a fraction over the same denominator (3x + 1). Then expand the top: (2x − 1)(3x + 1) = 6x² + 2x − 3x − 1 = 6x² − x − 1.
- Step 4: simplify. 6x² − 13x + 1−x + x = 0, so the x terms cancel. 6x² − 1 doesn’t factorise with (3x + 1), so this is the simplest form.
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