(7n - 4)(n + 3)
To factor a trinomial (three-term expression) of the form n2+bn+c, find the two numbers h and k that are factors of c and add up to b. Then, write those numbers in this template: (n+h)(n+k) For the trinomial n2+7n-44, c = -44 and b = 7. The two numbers that are factors of -44 and add up to 7 are 11 and -4. So, the factored form would be (n+11)(n-4).
22n + 2.
6m + 5n - 4m + 7n = 2m + 12n
(10, 2)
(n + 11)(n - 4)
-7n2 + 74n + 33 = -7n2 + 77n - 3n + 33 = -7n*(n - 11) - 3*(n - 11) = =(-7n - 3)*(n - 11) = -(7n +3)*(n - 11)
(7n - 4)(n + 3)
To factor a trinomial (three-term expression) of the form n2+bn+c, find the two numbers h and k that are factors of c and add up to b. Then, write those numbers in this template: (n+h)(n+k) For the trinomial n2+7n-44, c = -44 and b = 7. The two numbers that are factors of -44 and add up to 7 are 11 and -4. So, the factored form would be (n+11)(n-4).
22n + 2.
7n3 + 6n = n*(7n2 + 6)
n(2n - 1)(2n + 7)
6m + 5n - 4m + 7n = 2m + 12n
Gather like terms. The - 2 and 4; the 7n and 3n. 7n - 2 + 3n + 4 = 10n + 2
7n - 2 + 3n + 4 = 7n + 3n - 2 + 4 = 10n + 2 which can be simplified to 2(5n +1)
This question as written can be taken two ways: 9m + 6n - 3m + 7n = 6m + 13n, or (9m + 6n) - (3m + 7n) = 9m + 6n - 3m - 7n = 6m - n
10n + 3 = 7n + 53n = 2n = 2/3