How do you factor 14n^2+23n-1514n2+23n−15?
1 Answer
Explanation:
Given:
Use an AC method.
Find a pair of factors of
The pair
Use this pair to split the middle term and factor by grouping:
14n^2+23n-1514n2+23n−15
=14n^2+30n-7n-15=14n2+30n−7n−15
=(14n^2+30n)-(7n+15)=(14n2+30n)−(7n+15)
=2n(7n+15)-1(7n+15)=2n(7n+15)−1(7n+15)
=(2n-1)(7n+15)=(2n−1)(7n+15)
Alternatively, multiply by
a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)
with
56(14n^2+23n-15)56(14n2+23n−15)
=784n^2+1288n-840=784n2+1288n−840
=(28n)^2+2(28n)(23)-840=(28n)2+2(28n)(23)−840
=(28n+23)^2-23^2-840=(28n+23)2−232−840
=(28n+23)^2-529-840=(28n+23)2−529−840
=(28n+23)^2-1369=(28n+23)2−1369
=(28n+23)^2-37^2=(28n+23)2−372
=((28n+23)-37)((28n+23)+37)=((28n+23)−37)((28n+23)+37)
=(28n-14)(28n+60)=(28n−14)(28n+60)
=(14(2n-1))(4(7n+15))=(14(2n−1))(4(7n+15))
=56(2n-1)(7n+15)=56(2n−1)(7n+15)
So:
14n^2+23n-15 = (2n-1)(7n+15)14n2+23n−15=(2n−1)(7n+15)
