Factorise quadratics with other leading coefficients
key notes:
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
➡️ Look at the given quadratic:
2j2+7j+5
The product a . c is 10, so you need to find a pair of factors with a product of 10. The b term is 7, so you need to find a pair of factors with a sum of 7. Since the product is positive (10) and the sum is positive (7), you need both factors to be positive.
Make a list of the possible factor pairs with a product of 10, and then find the one with a sum of 7.
The factors 2 and 5 have a sum of 7. So, replace the quadratic’s 7j term with 2j and 5j, and then factor by grouping.
2j2+7j+5
2j2+2j+5j+5
2j(j+1)+5(j+1) Factor by grouping; the expressions in brackets should match
(2j+5)(j+1)
Finally, check your work.
(2j+5)(j+1)
2j2+5j+2j+5 Apply the distributive property (FOIL)
2j2+7j+5
Yes, 2j2+7j+5=(2j+5)(j+1).
➡️ Factorise.
3n2+10n+7
Look at the given quadratic:
3n2+10n+7
The product ac is 21, so you need to find a pair of factors with a product of 21. The b term is 10, so you need to find a pair of factors with a sum of 10. Since the product is positive (21) 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 21, and then find the one with a sum of 10.
The factors 3 and 7 have a sum of 10. So, replace the quadratic’s 10n term with 3n and 7n, and then factor by grouping.
3n2+10n+7
3n2+3n+7n+7
3n(n+1) + 7(n+1) Factor by grouping; the expressions in brackets should match
(3n+7)(n+1)
Finally, check your work.
(3n+7)(n+1)
3n2+7n+3n+7 Apply the distributive property (FOIL)
3n2+10n+7
Yes, 3n2+10n+7=(3n+7)(n+1).
➡️ Factorise.
2q2+11q+9
Look at the given quadratic:
2q2+11q+9
The product ac is 18, so you need to find a pair of factors with a product of 18. The b term is 11, so you need to find a pair of factors with a sum of 11. Since the product is positive (18) and the sum is positive (11), 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 11.
The factors 2 and 9 have a sum of 11. So, replace the quadratic’s 11q term with 2q and 9q, and then factor by grouping.
2q2+11q+9
2q2+2q+9q+9
2q(q+1)+9(q+1) Factor by grouping; the expressions in brackets should match
(2q+9)(q+1)
Finally, check your work.
(2q+9)(q+1)
2q2+9q+2q+9 Apply the distributive property (FOIL)
2q2+11q+9
Yes, 2q2+11q+9=(2q+9)(q+1).
let’s practice!
