Question

How do you factor `14n^2+23n-15`?

Answer 1

`14n^2+23n-15=(2n-1)(7n+15)`

Explanation:

Given: `14n^2+23n-15`

Use an AC method.

Find a pair of factors of `AC = 14*15=210` which differ by `B=23`

The pair `30, 7` works.

Use this pair to split the middle term and factor by grouping:

`14n^2+23n-15`

`=14n^2+30n-7n-15`

`=(14n^2+30n)-(7n+15)`

`=2n(7n+15)-1(7n+15)`

`=(2n-1)(7n+15)`

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Alternatively, multiply by `4*14 = 56`, make into a difference of squares, factor using the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

with `a=(28n+23)` and `b=37`, then divide by `56`:

`56(14n^2+23n-15)`

`=784n^2+1288n-840`

`=(28n)^2+2(28n)(23)-840`

`=(28n+23)^2-23^2-840`

`=(28n+23)^2-529-840`

`=(28n+23)^2-1369`

`=(28n+23)^2-37^2`

`=((28n+23)-37)((28n+23)+37)`

`=(28n-14)(28n+60)`

`=(14(2n-1))(4(7n+15))`

`=56(2n-1)(7n+15)`

So:

`14n^2+23n-15 = (2n-1)(7n+15)`