Question

How do you factor `18c^2-3c-10`?

Answer 1

`18c^2-3c-10=(6c-5)(3c+2)`

Explanation:

We can use an AC method:

Look for a pair of factors of `AC = 18*10 = 180`, with difference `B=3`.

The pair `12, 15` works.

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

`18c^2-3c-10`

`=18c^2+12c-15c-10`

`=(18c^2+12c)-(15c+10)`

`=6c(3c+2)-5(3c+2)`

`=(6c-5)(3c+2)`

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Alternative method

Alternatively, we can find this factorisation by completing the square then using the difference of squares identity:

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

with `a=(12c-1)` and `b=9` as follows:

To make the arithmetic easier, first multiply by `8`, remembering to divide by `8` at the end:

`8(18c^2-3c-10)`

`=144c^2-24c-80`

`=(12c^2)-2(12c)-80`

`=(12c-1)^2-1-80`

`=(12c-1)^2-81`

`=(12c-1)^2-9^2`

`=((12c-1)-9)((12c-1)+9)`

`=(12c-10)(12c+8)`

`=(2(6c-5))(4(3c+2))`

`=8(6c-5)(3c+2)`

So dividing both ends by `8` we find:

`18c^2-3c-10 = (6c-5)(3c+2)`

Why did I multiply by `8` first?

If I multiplied by `2` then that would make the leading term `36c^2 = (6c)^2`, but then the completed square would have involved `(6c-1/2)^2 = 36c^2-6c+1/4`. To avoid the messiness of fractions, multiply by an extra factor of `2^2 = 4` first.