Question

How do you factor `w^2-7w+12`?

Answer 1

`w^2-7w+12 =(w-3)(w-4)`

Explanation:

Assuming (hopefully) that `w^2-7w+12)` can be written as `(w-a)(w-b)` for some integer constants `a` and `b`
`color(white)("XXXX")` `color(white)("XXXX")`(Note:
`color(white)("XXXX")` `color(white)("XXXX")`we know the signs on `a` and `b` must be both negative
`color(white)("XXXX")` `color(white)("XXXX")`since the third term is positive
`color(white)("XXXX")` `color(white)("XXXX")`and the second term is negative)

We are looking for factors of 12 whose sum is 7:
`color(white)("XXXX")` `1xx12`:`color(white)("XXXX")` `1+12 != 7`
`color(white)("XXXX")` `2xx6`:`color(white)("XXXX")` `2+6 +=7`
`color(white)("XXXX")` `3xx4`:`color(white)("XXXX")` `3+4 = 7`
...and we've found our required values for `a` and `b`

Answer 2

Factor: `y = w^2 - 7w + 12`

Ans: (x - 3)(x - 4)

Explanation:

Find 2 number p and q knowing sum (-7) and product (12). p and q have same sign (Rule of sign)
Factor pairs of 12 --> (2, 6)(3, 4). This sum is 7 = -b. Then p = -3 and q = -4
Factored form: `y = (x - 3)(x - 4).`