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AQA GCSE · Question 20 · Algebra
Factorise fully 3x² + 23x + 30
Factorise fully 3x² + 23x + 30
How to approach this question
1. This is a quadratic expression in the form ax² + bx + c, where a=3, b=23, c=30.
2. Since a > 1, we look for two numbers that multiply to give a*c and add to give b.
3. Calculate a*c = 3 * 30 = 90.
4. We need two numbers that multiply to 90 and add to 23. Let's list factors of 90: (1,90), (2,45), (3,30), (5,18), (6,15), (9,10).
5. The pair that adds to 23 is 5 and 18.
6. Rewrite the middle term (23x) using these two numbers: 3x² + 18x + 5x + 30.
7. Factorise the first two terms and the last two terms separately: 3x(x + 6) + 5(x + 6).
8. Since the bracket (x + 6) is common, we can factor it out: (3x + 5)(x + 6).
Full Answer
(3x + 5)(x + 6)
To factorise the quadratic expression 3x² + 23x + 30, we can use the "ac method".
The expression is in the form ax² + bx + c, with a=3, b=23, and c=30.
Step 1: Multiply a and c.
a × c = 3 × 30 = 90.
Step 2: Find two numbers that multiply to give 90 and add to give b (which is 23).
Let's list the factor pairs of 90:
1 × 90
2 × 45
3 × 30
5 × 18 (5 + 18 = 23. This is the pair we need!)
6 × 15
9 × 10
Step 3: Split the middle term.
We rewrite 23x as 18x + 5x.
The expression becomes: 3x² + 18x + 5x + 30.
Step 4: Factorise by grouping.
Group the first two terms and the last two terms and find the common factor for each group.
(3x² + 18x) + (5x + 30)
Factor out 3x from the first group: 3x(x + 6)
Factor out 5 from the second group: 5(x + 6)
The expression is now: 3x(x + 6) + 5(x + 6).
Step 5: Final factorisation.
Both parts have a common factor of (x + 6). We can factor this out.
(3x + 5)(x + 6).
You can check the answer by expanding the brackets: (3x+5)(x+6) = 3x² + 18x + 5x + 30 = 3x² + 23x + 30.
Common mistakes
✗ Finding the correct numbers (18 and 5) but putting them directly into the brackets, e.g., (x+18)(x+5), which is incorrect for a non-monic quadratic.
✗ Making sign errors.
✗ Incorrectly factorising by grouping.
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