Factorise quadratics with leading coefficient1
Key Notes:
Understanding Quadratic Expressions
- A quadratic expression is in the form: x² + bx + c, where the leading coefficient (coefficient of x²) is 1.
Factorisation as Reverse Expansion
- Factorising means writing the quadratic expression as a product of two binomials.
- Example: x² + 5x + 6 = (x + 2)(x + 3).
Finding Two Numbers
Identify two numbers that:
- Multiply to give c (the constant term).
- Add to give b (the coefficient of x).
- Example: In x² + 7x + 12, find two numbers that multiply to 12 and add to 7 → (3 and 4).
Writing the Factors
- Once the numbers are found, express the quadratic as:
- x² + bx + c = (x + p)(x + q)
- Example: x² + 7x + 12 = (x + 3)(x + 4).
Checking the Factorisation
Expand the binomials to verify:
- (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✔
Special Cases
Perfect Square Quadratics: Example:
- x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)²
Difference of Squares: Example:
- x² – 9 = (x + 3)(x – 3).
To factorise a quadratic of the form ax2 + bx + c, write it as(x + r1) (x + r2)
where c = r1 . r2 and b = r1 + r2.
Learn with an example
👉 Factorise.
s2+5s+4=—–
Look at the given quadratic:
s2+5s+4
The c term is 4, so you need to find a pair of factors with a product of 4. The b term is 5, so you need to find a pair of factors with a sum of 5. Since the product is positive (4) and the sum is positive (5), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 4, and then find the one with a sum of 5.
| Factor pairs of c=4 | Sum of factor pairs |
| 1 . 4=4 | 1+4=5 |
| 2 . 2=4 | 2+2=4 |
The factors 1 and 4 have a sum of 5. Use those numbers to factorise s2+5s+4.
s2+5s+4
(s+1)(s+4)
Finally, check your work.
(s+1)(s+4)
s2+s+4s+4Apply the distributive property (FOIL)
s2+5s+4
Yes, s2+5s+4=(s+1)(s+4).
👉 Factorise.
f2+6f+8=—–
Look at the given quadratic:
f2+6f+8
The c term is 8, so you need to find a pair of factors with a product of 8. The b term is 6, so you need to find a pair of factors with a sum of 6. Since the product is positive (8) and the sum is positive (6), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 8, and then find the one with a sum of 6.
| Factor pairs of c=8 | Sum of factor pairs |
| 1 . 8=8 | 1+8=9 |
| 2 . 4=8 | 2+4=6 |
The factors 2 and 4 have a sum of 6. Use those numbers to factorise f2+6f+8.
f2+6f+8
(f+2)(f+4)
Finally, check your work.
(f+2)(f+4)
f2+2f+4f+8Apply the distributive property (FOIL)
f2+6f+8
Yes, f2+6f+8=(f+2)(f+4).
👉 Factorise.
x2+10x+9=——
Look at the given quadratic:
x2+10x+9
The c term is 9, so you need to find a pair of factors with a product of 9. The b term is 10, so you need to find a pair of factors with a sum of 10. Since the product is positive (9) and the sum is positive (10), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 9, and then find the one with a sum of 10.
| Factor pairs of c=9 | Sum of factor pairs |
| 1 . 9=9 | 1+9=10 |
| 3 . 3=9 | 3+3=6 |
The factors 1 and 9 have a sum of 10. Use those numbers to factorise x2+10x+9.
x2+10x+9
(x+1)(x+9)
Finally, check your work.
(x+1)(x+9)
x2+x+9x+9 Apply the distributive property (FOIL)
x2+10x+9
Yes, x2+10x+9=(x+1)(x+9).
let’s practice!
