Question

How do you factor the expression `25y^2- 52y + 27`?

Answer 1

`(25y-27)(y-1)`

Explanation:

We need to find the factors that when multiplied it gives `675` `(25 xx 27)` and when added it gives `-52`.

Multiplying:
`675` = `3xx3xx3xx5xx5` = `27 xx 25`

Adding:
`-52` = `-27-25`
Hence, `-27 xx -25 = 675`

So the factors are: `-27` and `-25`

`25y^2-52y+27`

`25y^2-25y-27y+27`

`25y(y-1)-27(y-1)`

`(25y-27)(y-1)`

Check the answer:
`(25y-27)(y-1)`
`(25yxxy)-25y-27y+27`
`25y^2-52y+27`

Answer 2

`25y^2-52y+27 = (y-1)(25y-27)`

Explanation:

Given:

`25y^2-52y+27`

Note that `25-52+27 = 0`

Hence `y=1` is a zero and `(y-1)` a factor.

The leading term of the other factor must be `25y` to get `25y^2` in the product and the trailing term must be `-27` in order to get `+27` in the product.

So we find:

`25y^2-52y+27 = (y-1)(25y-27)`