Question

How do you factor `2t^2+7t+3`?

Answer 1

Make `2t^(2) + 7t + 3` equal to zero to form a quadratic equation.

Here's the general form of the quadratic equation

`ax^(2) + bx + c =0`

You need to find two numbers that multiply to give `a * c` and that add to give `b`. For this example, the two numbers you need must multply to give `2 * 3 = 6` and add to give `"7"`.

To find them, just list all of the factors of `6` and try to find a pair that satisfies the criteria given

`6 = 1* 2 * 3`

In simple cases such as this one, these two numbers can be easily found to be `"6"` and `"1"`, since

`6 + 1 = 7` and `6 * 1 = 6`

So, your equation then becomes

`2t^(2) + 6t + t +3`

This then becomes

`(2t^(2) + 6t) + (t + 3) => 2t * (t + 3) + (t + 3)`, which can be written as

`(t + 3) * (2t + 1)` `->` this is how the original equation factors.