Factorise quadratics with other leading coefficients
key notes:
To factorize 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.
3m2+5m+2=———-
Look at the given quadratic:
3m2+5m+2
The product ac is 6, so you need to find a pair of factors with a product of 6. The b term is 5, so you need to find a pair of factors with a sum of 5. Since the product is positive (6) 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 6, and then find the one with a sum of 5.
The factors 2 and 3 have a sum of 5. So, replace the quadratic’s 5m term with 2m and 3m, and then factor by grouping.
3m2+5m+2
3m2+2m+3m+2
m(3m+2)+1(3m+2)Factor by grouping; the expressions in brackets should match
(m+1)(3m+2)
Finally, check your work.
(m+1)(3m+2)
3m2+3m+2m+2Apply the distributive property (FOIL)
3m2+5m+2
Yes, 3m2+5m+2=(m+1)(3m+2).
Factorise.
2t2+13t+11=———
Look at the given quadratic:
2t2+13t+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.
The factors 2 and 11 have a sum of 13. So, replace the quadratic’s 13t term with 2t and 11t, and then factor by grouping.
2t2+13t+11
2t2+2t+11t+11
2t(t+1)+11(t+1)Factor by grouping; the expressions in brackets should match
(2t+11)(t+1)
Finally, check your work.
(2t+11)(t+1)
2t2+11t+2t+11 Apply the distributive property (FOIL)
2t2+13t+11
Yes, 2t2+13t+11=(2t+11)(t+1).
Factorise.
2t2+9t+9=———
Look at the given quadratic:
2t2+9t+9
The product ac is 18, so you need to find a pair of factors with a product of 18. The b term is 9, so you need to find a pair of factors with a sum of 9. Since the product is positive (18) 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 18, and then find the one with a sum of 9.
The factors 3 and 6 have a sum of 9. So, replace the quadratic’s 9t term with 3t and 6t, and then factor by grouping.
2t2+9t+9
2t2+3t+6t+9
t(2t+3)+3(2t+3)Factor by grouping; the expressions in brackets should match
(t+3)(2t+3)
Finally, check your work.
(t+3)(2t+3)
2t2+6t+3t+9 Apply the distributive property (FOIL)
2t2+9t+9
Yes, 2t2+9t+9=(t+3)(2t+3).
let’s practice!
