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).
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n2 + 7n - 44 (n + 11)(n - 4)
38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...
n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
2(n2+3n-54) 2(n+9)(n-6)
3 MgCl2 + N2 = Mg3N2 + 3 Cl2