Factorise quadratics with other leading coefficients
Key Notes:
Understanding Quadratic Equations
- A quadratic equation is in the form ax² + bx + c, where a ≠ 1 (i.e., the coefficient of x² is not 1).
- The goal of factorization is to express the quadratic as a product of two binomials.
Factorization Process
- Step 1: Multiply a and c (the coefficient of x² and the constant term).
- Step 2: Find two numbers that multiply to a × c and add to b (the coefficient of x).
- Step 3: Rewrite the middle term using the two numbers found.
- Step 4: Group the terms in pairs.
- Step 5: Factor out the common term from each group.
- Step 6: Write the final factorized form.
Example 1: Factorize 2x² + 7x + 3
- Multiply a × c: 2 × 3 = 6
- Find two numbers that multiply to 6 and add to 7 → 6 and 1
- Rewrite: 2x² + 6x + 1x + 3
- Group: (2x² + 6x) + (1x + 3)
- Factor out common terms: 2x(x + 3) + 1(x + 3)
- Final factorized form: (2x + 1)(x + 3)
Example 2: Factorize 3x² – 8x – 3
- Multiply a × c: 3 × (-3) = -9
- Find two numbers that multiply to -9 and add to -8 → -9 and 1
- Rewrite: 3x² – 9x + 1x – 3
- Group: (3x² – 9x) + (1x – 3)
- Factor out common terms: 3x(x – 3) + 1(x – 3)
- Final factorized form: (3x + 1)(x – 3)
Special Cases
- Perfect Square Quadratics: If the quadratic is of the form a²x² + 2abx + b², it factors as (ax + b)².
- Difference of Squares: If the quadratic is a²x² – b², it factors as (ax – b)(ax + b).
Checking Your Answer
- Expand the factorized expression to verify if it equals the original quadratic.
- If incorrect, revisit the factor pairs and adjust the grouping.
To factorise a quadratic of the form ax2+bx+c, write it as
ax2+r1x+r2x+c
where a . c=r1 . r2 and b=r1+r2. Then factor by grouping.
Learn with an example
🏀 Factorise.
2f2+13f+11=——
Look at the given quadratic:
2f2+13f+11
The product ac is 22, so you need to find a pair of factors with a product of 22. The b term is 13, so you need to find a pair of factors with a sum of 13. Since the product is positive (22) and the sum is positive (13), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 22, and then find the one with a sum of 13.
| Factor pairs of ac=22 | Sum of factor pairs |
| 1 . 22=22 | 1+22=23 |
| 2 . 11=22 | 2+11=13 |
The factors 2 and 11 have a sum of 13. So, replace the quadratic’s 13f term with 2f and 11f, and then factor by grouping.
2f2+13f+11
2f2+2f+11f+11
2f(f+1)+11(f+1) Factor by grouping; the expressions in brackets should match
(2f+11)(f+1)
Finally, check your work.
(2f+11)(f+1)
2f2+11f+2f+11 Apply the distributive property (FOIL)
2f2+13f+11
Yes, 2f2+13f+11=(2f+11)(f+1).
🏀 Factorise.
2m2+13m+11=——
Look at the given quadratic:
2m2+13m+11
The product ac is 22, so you need to find a pair of factors with a product of 22. The b term is 13, so you need to find a pair of factors with a sum of 13. Since the product is positive (22) and the sum is positive (13), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 22, and then find the one with a sum of 13.
| Factor pairs of ac=22 | Sum of factor pairs |
| 1 . 22=22 | 1+22=23 |
| 2 . 11=22 | 2+11=13 |
The factors 2 and 11 have a sum of 13. So, replace the quadratic’s 13m term with 2m and 11m, and then factor by grouping.
2m2+13m+11
2m2+2m+11m+11
2m(m+1)+11(m+1) Factor by grouping; the expressions in brackets should match
(2m+11)(m+1)
Finally, check your work.
(2m+11)(m+1)
2m2+11m+2m+11 Apply the distributive property (FOIL)
2m2+13m+11
Yes, 2m2+13m+11=(2m+11)(m+1).
🏀 Factorise.
2u2+9u+10=——
Look at the given quadratic:
2u2+9u+10
The product ac is 20, so you need to find a pair of factors with a product of 20. The b term is 9, so you need to find a pair of factors with a sum of 9. Since the product is positive (20) and the sum is positive (9), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 20, and then find the one with a sum of 9.
| Factor pairs of ac=20 | Sum of factor pairs |
| 1 . 20=20 | 1+20=21 |
| 2 . 10=20 | 2+10=12 |
| 4 . 5=20 | 4+5=9 |
The factors 4 and 5 have a sum of 9. So, replace the quadratic’s 9u term with 4u and 5u, and then factor by grouping.
2u2+9u+10
2u2+4u+5u+10
2u(u+2)+5(u+2)
Factor by grouping; the expressions in brackets should match
(2u+5)(u+2)
Finally, check your work.
(2u+5)(u+2)
2u2+5u+4u+10
Apply the distributive property (FOIL)
2u2+9u+10
Yes, 2u2+9u+10=(2u+5)(u+2).
let’s practice!
