Question

How do you factor the trinomial `10t² + t - 2`?

Answer 1

`10t^2 + t - 2 = (2t + 1)(5t-2)`

Explanation:

A quick way of doing it which works in nice cases is to look for numbers `a, b, c, d` such that
`(at + b)(ct + d) = 10t^2 + t - 2`

that is,
`{(ac = 10), (ad + bc = 1), (bd = -2):}`

As this ends up being a nice case, we will be able to find integers satisfying the above requirements. In the more difficult cases we can use the quadratic formula (see below).

If we are only looking at integers, then we know from the first equation that the options for `a` and `c` are `1` and `10` or `2` and `5`.
Also, from the third equation, we know that `b` and `d` have the options `1` and `-2` or `-1` and `2`. Checking to see which (if any) of these work leads us to find that
`a = 2`
`b = 1`
`c = 5`
`d = -2`
fulfills all the needed equations. Thus

`10t^2 + t - 2 = (2t + 1)(5t-2)`

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The following is a general method for finding out how to factor a quadratic polynomial using the quadratic equation, and is not needed for understanding the above solution.

We start with the goal of factoring a quadratic equation, that is, finding real values `n_1` and `n_2` such that
`ax^2 + bx + c = (x+n_1)(x+n_2)`

To do this, we consider the equation
`(x+n_1)(x+n_2)=0`
and note that it has the solutions `x = -n_1` and `x = -n_2`.

Thus in order to find `n_1` and `n_2`, we need only find the solutions to `ax^2 + bx + c = 0` and multiply by `-1`.

The quadratic formula states that if `ax^2 + bx + c = 0` then

`x = (-b+-sqrt(b^2-4ac))/(2a)`

Thus, multiplying by `-1`, we get

`n_1 = (b - sqrt(b^2-4ac))/(2a)`
and
`n_2 = (b + sqrt(b^2-4ac))/(2a)`

Note that if `b^2-4ac = 0` then `n_1 = n_2 = b/(2a)` and so
`ax^2 + bx + c = (x + b/(2a))^2`

Also, if `b^2 < 0` then there are no real numbers `n_1` and `n_2` which fulfill the needed conditions, as it is not possible to evaluate the square root of a negative number as a real number.