Class 8 · 8th standard · 2026-27

Class 8 maths worksheets, free, with every step on the page

Three free Class 8 maths worksheets, with every question and every step printed on this page rather than behind a download. These are not an even sample of the syllabus and that is deliberate: identities, factorisation and linear equations are what Class 9 polynomials and Class 10 quadratic equations are built on, and neither year sets aside time to teach them again. Every exercise opens with one question worked through in full.

Book a free assessment

3 worksheets · 102 questions · every answer on the page · no sign-up · aligned to NCERT's Class 8 maths book is Mathematics. Schools also use their own texts, so these sheets are written to the ideas rather than to one book's chapter order.

What these practise

The three standard identities, and recognising them inside a question
Using an identity to do arithmetic mentally - 103 × 97 in one line
Rearranging an identity to find a² + b² from a + b and ab
Factorising by highest common factor
Difference of squares and perfect square trinomials
Factorising four terms by grouping
Splitting the middle term, including when the leading coefficient is not 1
Linear equations with the variable on both sides and with brackets
Clearing fractional coefficients by multiplying through by the LCM
Forming an equation from a described situation and checking against the words

Last updated 2026-07-29

Free, with answers

The worksheets themselves

Every question is on this page, and so is every answer. Nothing is behind a sign-up, and there is no file to download unless you want one - use the print button on any sheet to save it as a PDF.

practice · medium · 30 min

Algebraic identities

Class 8 Maths · Algebraic expressions and identities · 36 questions with answers

Download PDF

There are only three identities in Class 8 and memorising them takes ten minutes. What the paper tests is recognition - seeing that x² - 49 is the third one in disguise, or that 103 × 97 can be done in one line. That recognition only comes from doing enough of them, which is why this is the chapter where practice genuinely cannot be substituted.

Expands using each of the three standard identities
Uses an identity to do arithmetic mentally
Recognises an identity inside a question that does not announce it
Rearranges an identity to find an unknown expression

Exercise 1 of 3 · 12 questions

Expand using an identity. Name the one you used.

Write the answer · Warm-up

Here’s one done for you

(2x + 3)² = 4x² + 12x + 9. The trap is writing 2x² instead of 4x² - the whole of 2x is being squared, not just the x. Whenever the first term has a number in front of it, square the number too. That single slip accounts for more lost marks in this chapter than any misremembered identity.

1)(x + 5)²
2)(a - 7)²
3)(2x + 3)²
4)(3m - 4)²
5)(x + 6)(x - 6)
6)(2y + 5)(2y - 5)
7)(x + 9)²
8)(a - 5)²
9)(3x + 7)²
10)(5m - 2)²
11)(x + 8)(x - 8)
12)(3y + 4)(3y - 4)

Answers

1) x² + 10x + 25, using (a + b)² = a² + 2ab + b².
2) a² - 14a + 49, using (a - b)² = a² - 2ab + b².
3) 4x² + 12x + 9. Here a = 2x, so a² = 4x² and 2ab = 2 × 2x × 3 = 12x.
4) 9m² - 24m + 16. Here a = 3m, so a² = 9m² and 2ab = 2 × 3m × 4 = 24m.
5) x² - 36, using (a + b)(a - b) = a² - b².
6) 4y² - 25. Here a = 2y, so a² = 4y².
7) x² + 18x + 81. Identity (a + b)² = a² + 2ab + b².
8) a² - 10a + 25. Identity (a - b)² = a² - 2ab + b².
9) 9x² + 42x + 49. Identity (a + b)², with a = 3x and b = 7.
10) 25m² - 20m + 4. Identity (a - b)², with a = 5m and b = 2.
11) x² - 64. Identity (a + b)(a - b) = a² - b².
12) 9y² - 16. Identity (a + b)(a - b), with a = 3y and b = 4.

Exercise 2 of 3 · 12 questions

Work these out using an identity. No long multiplication.

Write the answer · Guided

Here’s one done for you

103 × 97 = 9991, because 103 = 100 + 3 and 97 = 100 - 3, so the answer is 100² - 3² = 10000 - 9. This exercise exists for one reason: it turns three abstract formulas into something obviously useful. A child who can do 103 × 97 in their head stops treating identities as things to be memorised for a test.

1)102²
2)98²
3)103 × 97
4)51²
5)995 × 1005
6)49 × 51
7)101²
8)97²
9)104 × 96
10)52²
11)998 × 1002
12)48 × 52

Answers

1) 10404. (100 + 2)² = 10000 + 400 + 4.
2) 9604. (100 - 2)² = 10000 - 400 + 4.
3) 9991. (100 + 3)(100 - 3) = 10000 - 9.
4) 2601. (50 + 1)² = 2500 + 100 + 1.
5) 999975. (1000 - 5)(1000 + 5) = 1000000 - 25.
6) 2499. (50 - 1)(50 + 1) = 2500 - 1.
7) 10,201. (100 + 1)² = 10000 + 200 + 1.
8) 9,409. (100 - 3)² = 10000 - 600 + 9.
9) 9,984. (100 + 4)(100 - 4) = 10000 - 16.
10) 2,704. (50 + 2)² = 2500 + 200 + 4.
11) 999,996. (1000 - 2)(1000 + 2) = 1000000 - 4.
12) 2,496. (50 - 2)(50 + 2) = 2500 - 4.

Exercise 3 of 3 · 12 questions

Expand or evaluate. Some need an identity, some need rearranging.

Write the answer · Practice

Here’s one done for you

If a + b = 7 and ab = 12, then a² + b² = 25. You cannot find a and b separately from this - and you do not need to. Square the sum: (a + b)² = a² + 2ab + b², so 49 = a² + b² + 24, giving 25. This is the question type that separates memorising the identity from understanding it, and it appears in some form in every paper from here to Class 10.

1)(x + 3)(x + 5)
2)(x - 4)(x + 7)
3)(2a + 3)(a - 5)
4)If a + b = 7 and ab = 12, find a² + b².
5)If x + 1/x = 5, find x² + 1/x².
6)Simplify (x + y)² - (x - y)².
7)(x + 2)(x + 9)
8)(x - 3)(x + 8)
9)(3a + 2)(a - 4)
10)If a + b = 9 and ab = 20, find a² + b².
11)If x + 1/x = 6, find x² + 1/x².
12)Simplify (x + y)² + (x - y)².

Answers

1) x² + 8x + 15.
2) x² + 3x - 28.
3) 2a² - 7a - 15. Expanding gives 2a² - 10a + 3a - 15.
4) 25. Since (a + b)² = a² + 2ab + b², we get a² + b² = 49 - 2(12) = 25.
5) 23. Squaring both sides gives x² + 2 + 1/x² = 25, so x² + 1/x² = 23.
6) 4xy. The first expands to x² + 2xy + y² and the second to x² - 2xy + y²; subtracting leaves 4xy.
7) x² + 11x + 18.
8) x² + 5x - 24.
9) 3a² - 10a - 8.
10) a² + b² = (a + b)² - 2ab = 81 - 40 = 41.
11) x² + 1/x² = (x + 1/x)² - 2 = 36 - 2 = 34.
12) 2x² + 2y². The cross terms cancel because one is +2xy and the other -2xy.

While your child works

When the first term has a number in front - (2x + 3)² - check they squared the number too. That single slip is the commonest error in the chapter.
The mental arithmetic exercise is the one that makes identities feel worth knowing. Do not skip it if time is short.
The last two questions cannot be done by memorising. If those come out, the chapter is genuinely understood.

Open the Algebraic expressions and identities worksheet on its own page

practice · challenging · 30 min

Factorisation

Class 8 Maths · Factorisation · 34 questions with answers

Download PDF

Most children who are stuck on factorisation are not unable to factorise - they are guessing which method to try. There are four, and they go in a fixed order: common factor, identity, grouping, splitting the middle term. A child working down that list rarely gets stuck; a child starting anywhere gets stuck constantly.

Takes out the highest common factor before anything else
Recognises a difference of squares and a perfect square trinomial
Factorises a four-term expression by grouping
Splits the middle term of a quadratic, including when the leading coefficient is not 1
Chooses the method in the right order rather than by trial

Exercise 1 of 3 · 10 questions

Factorise by taking out the common factor.

Write the answer · Warm-up

Here’s one done for you

12a²b - 18ab² = 6ab(2a - 3b). Two things to check every time: the number factor is the highest common one (6, not 2 or 3), and the letters are taken to the lowest power that appears in both (ab, not a²b). Taking out a partial factor is not wrong exactly, but it leaves work behind and loses marks that were already earned.

1)6x + 9
2)12a²b - 18ab²
3)5x² + 10x
4)7pq + 14p
5)4m³ - 8m²
6)8x + 20
7)15a²b - 25ab²
8)6x² + 9x
9)3m³ - 12m²
10)9pq + 12p

Answers

1) 3(2x + 3).
2) 6ab(2a - 3b). The common factor is 6ab - take the highest one, not just any one.
3) 5x(x + 2).
4) 7p(q + 2).
5) 4m²(m - 2).
6) 4(2x + 5)
7) 5ab(3a - 5b)
8) 3x(2x + 3)
9) 3m²(m - 4)
10) 3p(3q + 4)

Exercise 2 of 3 · 12 questions

Factorise using an identity, or by grouping.

Write the answer · Guided

Here’s one done for you

ax + ay + bx + by = (a + b)(x + y). Grouping works when there are four terms and no factor common to all of them. Take a from the first pair and b from the second: a(x + y) + b(x + y). Both now share (x + y), so it comes out. If the second bracket does not match the first, try pairing the terms differently before deciding grouping will not work.

1)x² - 16
2)9a² - 25
3)x² + 10x + 25
4)4y² - 12y + 9
5)ax + ay + bx + by
6)x² + 3x + 2x + 6
7)x² - 49
8)16a² - 81
9)x² + 14x + 49
10)9y² - 24y + 16
11)px + py + qx + qy
12)x² + 5x + 3x + 15

Answers

1) (x + 4)(x - 4). Difference of squares.
2) (3a + 5)(3a - 5). Since 9a² = (3a)².
3) (x + 5)². A perfect square, because 25 = 5² and 10x = 2 × x × 5.
4) (2y - 3)². Since 4y² = (2y)², 9 = 3² and 12y = 2 × 2y × 3.
5) (a + b)(x + y). Group as a(x + y) + b(x + y).
6) (x + 3)(x + 2). Group as x(x + 3) + 2(x + 3).
7) (x + 7)(x - 7)
8) (4a + 9)(4a - 9)
9) (x + 7)²
10) (3y - 4)²
11) (x + y)(p + q)
12) (x + 3)(x + 5)

Exercise 3 of 3 · 12 questions

Factorise by splitting the middle term.

Write the answer · Practice

Here’s one done for you

2x² + 7x + 3 = (2x + 1)(x + 3). When the number in front of x² is not 1, the two numbers must multiply to the first coefficient times the last - here 2 × 3 = 6 - and add to the middle one, 7. That gives 6 and 1, so write 2x² + 6x + x + 3 and group: 2x(x + 3) + 1(x + 3). Children who learn the shortcut for x² + bx + c and are never shown this extension get stuck on every question where the leading coefficient is not 1.

1)x² + 7x + 12
2)x² - 5x + 6
3)x² + x - 12
4)x² - 2x - 15
5)2x² + 7x + 3
6)3x² - 10x + 8
7)x² + 11x + 30
8)x² - x - 20
9)x² - 9x + 20
10)x² + 2x - 24
11)2x² + 9x + 4
12)3x² + 11x + 6

Answers

1) (x + 3)(x + 4). Two numbers multiplying to 12 and adding to 7.
2) (x - 2)(x - 3). Product 6, sum -5, so both are negative.
3) (x + 4)(x - 3). Product -12, sum +1.
4) (x - 5)(x + 3). Product -15, sum -2.
5) (2x + 1)(x + 3). Product 2 × 3 = 6, sum 7, so split as 6x + x: 2x² + 6x + x + 3.
6) (3x - 4)(x - 2). Product 3 × 8 = 24, sum -10, so split as -6x - 4x.
7) (x + 5)(x + 6)
8) (x - 5)(x + 4)
9) (x - 4)(x - 5)
10) (x + 6)(x - 4)
11) (2x + 1)(x + 4)
12) (3x + 2)(x + 3)

While your child works

Ask 'which method, and why that one?' before any factorising happens. Common factor, identity, grouping, middle term - in that order.
Check the answer by expanding it back. It takes fifteen seconds and catches every error.
When the number in front of x² is not 1, the product to aim for is first coefficient times last. That extension is the one most often never taught.

Open the Factorisation worksheet on its own page

practice · medium · 30 min

Linear equations in one variable

Class 8 Maths · Linear equations · 32 questions with answers

Download PDF

Class 8 equations are the Class 7 ones with fractions added and a good deal less patience. The method has not changed - do the same thing to both sides - but a child who was guessing rather than solving cannot hide it once denominators appear. Every answer here is checkable by substitution, and checking is half the skill.

Solves an equation with the variable on both sides
Clears fractions by multiplying through by the LCM
Expands brackets correctly before collecting terms
Forms an equation from a described situation
Verifies an answer by substituting it back

Exercise 1 of 3 · 10 questions

Solve. Show each step.

Write the answer · Warm-up

Here’s one done for you

7 - 3x = x - 5 gives x = 3. Adding 3x to both sides rather than subtracting x is the easier route, because it leaves the x term positive - 7 = 4x - 5. Both routes reach the same answer; one of them produces far fewer sign errors, and choosing it deliberately is a habit worth building now.

1)5x - 3 = 12
2)2x + 7 = 3x - 4
3)3(x - 2) = 12
4)4x + 5 = 2x + 13
5)7 - 3x = x - 5
6)6x + 4 = 22
7)5x - 9 = 2x + 6
8)4(x + 3) = 20
9)9 - 2x = x + 3
10)7x - 5 = 4x + 10

Answers

1) x = 3. Add 3 to both sides to get 5x = 15, then divide by 5.
2) x = 11. Subtract 2x from both sides to get 7 = x - 4, then add 4.
3) x = 6. Divide both sides by 3 to get x - 2 = 4, then add 2.
4) x = 4. Subtract 2x to get 2x + 5 = 13, subtract 5, then divide by 2.
5) x = 3. Add 3x to both sides to get 7 = 4x - 5, add 5 to get 12 = 4x.
6) x = 3. Subtract 4: 6x = 18. Divide by 6.
7) x = 5. Subtract 2x: 3x - 9 = 6. Add 9: 3x = 15. Divide by 3.
8) x = 2. Divide by 4: x + 3 = 5. Subtract 3.
9) x = 2. Add 2x: 9 = 3x + 3. Subtract 3: 3x = 6. Divide by 3.
10) x = 5. Subtract 4x: 3x - 5 = 10. Add 5: 3x = 15. Divide by 3.

Exercise 2 of 3 · 10 questions

Solve. Clear the fractions first.

Write the answer · Guided

Here’s one done for you

x/2 + x/3 = 5 gives x = 6. Multiply every term by 6, the LCM of the denominators, and the fractions vanish: 3x + 2x = 30. The word to insist on is <em>every</em> - multiplying the left side and forgetting the 5 on the right is the single commonest error in this exercise, and the answer it produces looks entirely reasonable.

1)x/2 + x/3 = 5
2)(2x - 1)/3 = 5
3)x/4 - 2 = 1
4)(x + 1)/2 = (x - 1)/3
5)(3x + 2)/5 = (2x - 3)/3
6)x/3 + x/4 = 7
7)(3x - 2)/5 = 4
8)x/6 - 3 = 1
9)(x + 2)/3 = (x - 2)/5
10)(2x + 1)/3 = (x + 4)/2

Answers

1) x = 6. Multiply through by 6: 3x + 2x = 30, so 5x = 30.
2) x = 8. Multiply both sides by 3: 2x - 1 = 15.
3) x = 12. Add 2 to get x/4 = 3, then multiply by 4.
4) x = -5. Cross-multiplying gives 3(x + 1) = 2(x - 1), so 3x + 3 = 2x - 2.
5) x = 21. Cross-multiplying gives 3(3x + 2) = 5(2x - 3), so 9x + 6 = 10x - 15.
6) x = 12. Multiply throughout by 12: 4x + 3x = 84, so 7x = 84.
7) x = 22/3. Multiply both sides by 5: 3x - 2 = 20, so 3x = 22.
8) x = 24. Add 3: x/6 = 4. Multiply by 6.
9) x = -8. Cross-multiply: 5(x + 2) = 3(x - 2), so 5x + 10 = 3x - 6 and 2x = -16.
10) x = 10. Cross-multiply: 2(2x + 1) = 3(x + 4), so 4x + 2 = 3x + 12.

Exercise 3 of 3 · 12 questions

Form an equation and solve it. Check your answer against the question.

Write the answer · Practice

Here’s one done for you

The ages: let the son be s, so the father is 3s. In twelve years they are s + 12 and 3s + 12, and the father is then twice the son: 3s + 12 = 2(s + 12). That gives s = 12, so the son is 12 and the father 36. Check it against the question - in twelve years they are 24 and 48, and 48 is twice 24. Age problems are where children most often solve correctly and answer the wrong question, so checking against the words rather than the equation is what secures the mark.

1)The sum of three consecutive whole numbers is 51. Find them.
2)The length of a rectangle is twice its breadth and its perimeter is 48 cm. Find both.
3)A number increased by 8 gives three times the original number. Find it.
4)A father is three times as old as his son. In 12 years he will be twice as old. Find both ages.
5)Two numbers are in the ratio 3 : 5 and add up to 64. Find them.
6)The denominator of a fraction is 4 more than the numerator. Adding 1 to both gives 3/4. Find the fraction.
7)The sum of two consecutive odd numbers is 36. Find them.
8)The perimeter of a rectangle is 44 cm and the length is 4 cm more than the breadth. Find both.
9)A number decreased by 6 gives half the number. Find it.
10)A mother is four times as old as her son; in 6 years she will be three times as old. Find both ages.
11)Two numbers are in the ratio 2 : 7 and their sum is 63. Find them.
12)The numerator of a fraction is 3 less than the denominator, and the fraction equals 4/5. Find it.

Answers

1) n + (n + 1) + (n + 2) = 51, so 3n + 3 = 51 and n = 16. The numbers are 16, 17 and 18.
2) 2(2b + b) = 48, so 6b = 48 and b = 8. The breadth is 8 cm and the length is 16 cm.
3) n + 8 = 3n, so 2n = 8 and n = 4.
4) Let the son be s. Then 3s + 12 = 2(s + 12), so 3s + 12 = 2s + 24 and s = 12. The son is 12 and the father is 36.
5) 3x + 5x = 64, so 8x = 64 and x = 8. The numbers are 24 and 40.
6) Let the numerator be n. Then (n + 1)/(n + 5) = 3/4, so 4n + 4 = 3n + 15 and n = 11. The fraction is 11/15.
7) n + (n + 2) = 36, so 2n + 2 = 36, 2n = 34 and n = 17. The numbers are 17 and 19.
8) 2(b + 4 + b) = 44, so 4b + 8 = 44, 4b = 36 and b = 9. The breadth is 9 cm and the length 13 cm.
9) n - 6 = n/2, so 2n - 12 = n and n = 12.
10) Let the son be s, so the mother is 4s. Then 4s + 6 = 3(s + 6), giving 4s + 6 = 3s + 18 and s = 12. The son is 12 and the mother 48.
11) 2k + 7k = 63, so 9k = 63 and k = 7. The numbers are 14 and 49.
12) Let the denominator be d, so the numerator is d - 3. Then (d - 3)/d = 4/5, giving 5d - 15 = 4d and d = 15. The fraction is 12/15.

While your child works

When fractions are cleared, check every term was multiplied - including the number on the right. That omission is the most frequent error here.
Substituting the answer back takes ten seconds and catches nearly every slip.
In word problems, check the answer against the question rather than against the equation. Age problems in particular are often solved correctly and reported wrongly.

Open the Linear equations worksheet on its own page

Why only these three chapters?

Because they are the ones Class 9 assumes and never reteaches.

Class 9 opens with polynomials, which are unusable without fluent expansion and factorisation. Class 10 quadratic equations are the same skill again. Meanwhile mensuration, data handling and quadrilaterals - all perfectly reasonable Class 8 chapters - are revisited later with fresh teaching.

So a family with two hours a week should not spread them evenly. That is the argument set out in full on our Class 8 maths tuition page, and these sheets follow it.

How should we use these at home?

Thirty minutes each, once or twice a week. Every exercise opens with one question worked through in full, because in all three chapters the method is what is being practised.

Ask 'which method, and why that one?' before any factorising. Common factor, identity, grouping, middle term - in that order.
Check by expanding back. Fifteen seconds, and it catches every factorisation error.
Substitute the answer back in every equation, then check it against the words of the question.
Let your child mark their own work. Every answer is on the page for exactly that reason.

Where do Class 8 children usually stall?

Three places, and each has a one-line fix.

Squaring only the letter. (2x + 3)² written as 2x² + 12x + 9 rather than 4x² + 12x + 9. The whole of 2x is being squared.
Guessing the factorisation method. Four methods, one order. A child working down the list rarely gets stuck; a child starting anywhere gets stuck constantly.
Multiplying only one side when clearing fractions. Every term, including the whole number on the right.

None of these is about mathematical ability. All three are habits, which is why Class 8 responds so quickly - and why the same gaps left until Class 9 cost a term rather than a fortnight.

Questions parents ask

Class 8 Maths worksheets - your questions

Are these Class 8 maths worksheets free?

Yes. Every question and every answer is on this page, with no sign-up, no email wall and no account. The printable version puts the answers on a separate final sheet.

Why only three chapters rather than the whole syllabus?

Because these three are what Class 9 and Class 10 algebra are built on, and neither year reteaches them. Mensuration and data handling are revisited later with fresh teaching; identities and factorisation are not.

How many identities are there in Class 8?

Three: (a + b)² = a² + 2ab + b², (a - b)² = a² - 2ab + b², and (a + b)(a - b) = a² - b². Memorising them is easy; recognising them inside a question is what the paper tests.

In what order should factorisation methods be tried?

Common factor first, then an identity, then grouping, then splitting the middle term. Most children who appear stuck are guessing which to use rather than unable to use any of them.

Why does my child struggle when the coefficient of x² is not 1?

Because the shortcut they learned only works when it is 1. For 2x² + 7x + 3 the two numbers must multiply to 2 × 3 = 6 and add to 7 - an extension that is very often never taught.

Are these matched to NCERT?

They are written to the ideas Class 8 maths tests rather than to one book's chapter order, so they work whether your school uses NCERT or its own text.

How long should a Class 8 child spend on one sheet?

About thirty minutes. If it runs much longer, work through the worked examples together first - the difficulty is usually the method rather than the arithmetic.

Can I print these worksheets?

Yes. They print cleanly on A4 with the answers on a separate final page, and nothing needs to be downloaded or signed up for.

100% free · no card

Worksheets are free. So is the first class.

If these showed up a gap in your child's maths, a teacher who has taught Class 8 can work through it with them. One free class, nothing to cancel.

You tell us the class, board and what is going wrong.
We match a teacher who has taught exactly that, and send you a time.
A free 20-minute assessment. Sit beside your child if you like.

Book a free assessment See class plans & fees

No card, no commitment
Live classes, never a crowd
The same teacher every week

Class 8 maths worksheets, free, with every step on the page

The worksheets themselves

Algebraic identities

Factorisation

Linear equations in one variable

Class 8 Maths - the guide

Why only these three chapters?

How should we use these at home?

Where do Class 8 children usually stall?

Class 8 Maths worksheets - your questions

Worksheets are free. So is the first class.

Book a free assessment