factor each completely and how you got the answer
2n^2 + 3n - 9
5n^2 + 19n + 12
2n^2 + 5n + 2
To factor each polynomial completely, we can use a combination of techniques such as factoring by grouping, trial and error, or the quadratic formula.
1. 2n^2 + 3n - 9:
This expression is in the form of a quadratic trinomial. To factor it, we can look for two numbers that multiply to the product of the coefficient of n^2 term and the constant term, and add up to the coefficient of the n term. In this case, the numbers are 2 and -9:
2n^2 + 6n - 3n - 9
2n(n + 3) - 3(n + 3)
(2n - 3)(n + 3)
2. 5n^2 + 19n + 12:
For this expression, we can also factor it by looking for two numbers that multiply to the product of the coefficient of n^2 term and the constant term, and add up to the coefficient of the n term. The numbers are 4 and 3:
5n^2 + 15n + 4n + 12
5n(n + 3) + 4(n + 3)
(5n + 4)(n + 3)
3. 2n^2 + 5n + 2:
Similar to the previous expressions, we need to find two numbers that multiply to the product of the coefficient of n^2 term and the constant term, and add up to the coefficient of the n term. The numbers are 2 and 1:
2n^2 + 4n + n + 2
2n(n + 2) + 1(n + 2)
(2n + 1)(n + 2)
Therefore, the factored forms of the given expressions are:
1. (2n - 3)(n + 3)
2. (5n + 4)(n + 3)
3. (2n + 1)(n + 2)