Question

Can `y= x^2-3x-28` be factored? If so what are the factors ?

Answer 1

`x^2 - 3x - 28 = (x-7)(x+4)`

Explanation:

Luckily, your `x^2` term already has the factor `1` in front - this makes your computation easier.

I would like to suggest the following method:

If your factorization exists, it would look like this:

`x^2 - 3x - 28 = (x + a)(x + b)`

`color(white)(xxxxxxxxx) = x^2 + (a+b)x + a*b`

So, your goal is to find `a` and `b` with the following properties:

• `a + b = -3`
• `color(white)(i)a xx b color(white)(i) = -28`

It is always easier to start with the multiplication. Since `-28` is negative, one of the two numbers needs to be positive and the other one negative.
Possible integer pairs of `a` and `b` are:

• `(-4) xx 7 color(white)(xx)` or `color(white)(xx) 4 xx (-7)`
• `(-2) xx 14 color(white)(xii)` or `color(white)(xx)2 xx (-14)`
• `(-1) xx 28 color(white)(xii)` or `color(white)(xx)1 xx (-28)`

It is easy to check that the correct pair is `4` and `-7` since `4 + (- 7) = -3` and `4 * (-7) = -28`.

Thus, your factorization is

`x^2 - 3x - 28 = (x-7)(x+4)`.

`color(white)(x)`

By the way, this also means that the solutions for the equation

`x^2 - 3x - 28 = 0`

are `-a` and `-b`, so `x = 7` or `x = -4`.