Question

Factorise the following:

3(x – 2y)² + 4(x – 2y) – 15

Answer 1

We are given the expression:

3(x – 2y)² + 4(x – 2y) – 15

Step 1: Substitution for simplification

Let z = (x – 2y).

Substituting this into the expression simplifies it to 3z² + 4z – 15.

Step 2: Factor the quadratic expression

Now, we need to factor 3z² + 4z – 15.

We are looking for two numbers that multiply to 3 × (–15) = –45 (the product of the coefficient of z² and the constant term) and add up to 4 (the coefficient of z).

The two numbers that satisfy this are 9 and –5, because:

9 × (–5) = –45 and 9 + (–5) = 4

Step 3: Rewrite the middle term

We can now rewrite 4z as 9z – 5z:

3z² + 9z – 5z – 15

Step 4: Group the terms

Now, group the terms in pairs:

(3z² + 9z) – (5z + 15)

Step 5: Factor each group

Factor out the common factor from each group:

3z(z + 3) – 5(z + 3)

Step 6: Factor out the common binomial

Now, factor out the common binomial factor (z + 3):

(3z – 5) (z + 3)

Step 7: Substitute back for z

Now, substitute z = (x – 2y) back into the factored form:

[3(x – 2y) – 5] [(x – 2y) + 3]

Final factorisation:

Thus, the fully factorised form is (3x – 6y – 5) (x – 2y + 3)