Math worksheet featuring a variety of factorisation problems for students to solve.

Let's solve each part of the problem step by step.

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1. Factorise:

We are to factorise linear expressions by taking out the highest common factor (HCF) from each term.

#### (a) $ 2x + 4 $

- HCF of 2 and 4 is 2
- $ 2x + 4 = 2(x + 2) $

✔ Answer: $ \boxed{2(x + 2)} $

#### (b) $ 5x + 15 $

- HCF of 5 and 15 is 5
- $ 5x + 15 = 5(x + 3) $

✔ Answer: $ \boxed{5(x + 3)} $

#### (c) $ 6x + 18 $

- HCF of 6 and 18 is 6
- $ 6x + 18 = 6(x + 3) $

✔ Answer: $ \boxed{6(x + 3)} $

#### (d) $ 5x - 25 $

- HCF of 5 and 25 is 5
- $ 5x - 25 = 5(x - 5) $

✔ Answer: $ \boxed{5(x - 5)} $

#### (e) $ 3x - 21 $

- HCF of 3 and 21 is 3
- $ 3x - 21 = 3(x - 7) $

✔ Answer: $ \boxed{3(x - 7)} $

#### (f) $ 7x + 35 $

- HCF of 7 and 35 is 7
- $ 7x + 35 = 7(x + 5) $

✔ Answer: $ \boxed{7(x + 5)} $

#### (g) $ 9x - 12 $

- HCF of 9 and 12 is 3
- $ 9x - 12 = 3(3x - 4) $

✔ Answer: $ \boxed{3(3x - 4)} $

#### (h) $ 15x + 20 $

- HCF of 15 and 20 is 5
- $ 15x + 20 = 5(3x + 4) $

✔ Answer: $ \boxed{5(3x + 4)} $

#### (i) $ 42x + 15 $

- HCF of 42 and 15 is 3
- $ 42x + 15 = 3(14x + 5) $

✔ Answer: $ \boxed{3(14x + 5)} $

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2. Factorise:

Now we factorise expressions with quadratic terms, again by taking out the common factor.

#### (a) $ 3x^2 + x $

- Both terms have a factor of x
- $ 3x^2 + x = x(3x + 1) $

✔ Answer: $ \boxed{x(3x + 1)} $

#### (b) $ 5x^2 + 10 $

- HCF of 5 and 10 is 5
- $ 5x^2 + 10 = 5(x^2 + 2) $

✔ Answer: $ \boxed{5(x^2 + 2)} $

#### (c) $ 6x - 3x^2 $

- Rearranged: $ -3x^2 + 6x $
- HCF of 3x and 6x is 3x
- $ 6x - 3x^2 = 3x(2 - x) $

✔ Answer: $ \boxed{3x(2 - x)} $
*(Alternatively, you could write $ -3x(x - 2) $ — both are correct, but $ 3x(2 - x) $ is simpler here.)*

#### (d) $ 6x^2 - 4x $

- HCF of 6 and 4 is 2, and both terms have x
- So HCF is 2x
- $ 6x^2 - 4x = 2x(3x - 2) $

✔ Answer: $ \boxed{2x(3x - 2)} $

#### (e) $ 21x^2 + 14x $

- HCF of 21 and 14 is 7, and both have x
- HCF = 7x
- $ 21x^2 + 14x = 7x(3x + 2) $

✔ Answer: $ \boxed{7x(3x + 2)} $

#### (f) $ 15x - 25x^2 $

- Rearranged: $ -25x^2 + 15x $
- HCF of 15 and 25 is 5, and both have x
- HCF = 5x
- $ 15x - 25x^2 = 5x(3 - 5x) $

✔ Answer: $ \boxed{5x(3 - 5x)} $

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3. Denise states that:

$$
4x + 6x^2 = x(4 + 6x)
$$

#### (a) Is her statement true?

Let’s check:

- Right-hand side: $ x(4 + 6x) = 4x + 6x^2 $
- Left-hand side: $ 4x + 6x^2 $

So yes, algebraically, they are equal.

✔ So her statement is true.

But let’s think deeper — is it fully factored? Or could it be improved?

Yes, because both terms on the right have a common factor of 2, so we can factor further.

But Denise's expression is still correct, just not fully simplified.

So answer to (a):
✔ Yes, her statement is true.

#### (b) Describe how it could be improved.

The expression $ x(4 + 6x) $ has a common factor of 2 in the parentheses.

We can factor out the 2:

$$
x(4 + 6x) = x \cdot 2(2 + 3x) = 2x(2 + 3x)
$$

So the fully factored form is $ 2x(2 + 3x) $

✔ Improvement: Factor out the greatest common factor completely.

So, Denise’s expression is correct, but it can be improved by factoring out the GCF of all terms, which is 2x.

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✔ Final Answers:

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1. Factorise:

(a) $ 2(x + 2) $
(b) $ 5(x + 3) $
(c) $ 6(x + 3) $
(d) $ 5(x - 5) $
(e) $ 3(x - 7) $
(f) $ 7(x + 5) $
(g) $ 3(3x - 4) $
(h) $ 5(3x + 4) $
(i) $ 3(14x + 5) $

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2. Factorise:

(a) $ x(3x + 1) $
(b) $ 5(x^2 + 2) $
(c) $ 3x(2 - x) $
(d) $ 2x(3x - 2) $
(e) $ 7x(3x + 2) $
(f) $ 5x(3 - 5x) $

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3. Denise's statement:

(a) Yes, her statement is true.
(b) It could be improved by factoring out the greatest common factor completely. The full factorisation is:

$$
4x + 6x^2 = 2x(2 + 3x)
$$

So, while $ x(4 + 6x) $ is correct, the improved version is $ \boxed{2x(2 + 3x)} $.