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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).

Q: How do you factor n2 plus 7n - 44?

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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)

n2 + 9n + 18 = (n + 6)(n + 3)

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n2 + 7n - 44 (n + 11)(n - 4)

(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, ...

5(n2 + 2n + 4)

Yes because its discriminant is greater than zero.

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)

-mn(m - n2)(m + n2)

The formula for the synthesis of ammonia from diatomic nitrogen and hydrogen is: N2+3H2-->2NH3

n2 + n2 = 2 n2

.04m2-n2=(.2m+n)(.2m-n)

n2 - 4 is the "difference of two squares." That factors to (n - 2)(n+2)a2 + 2a + 1 is "additive squares." That factors to (a + 1)(a + 1) or (a + 1)2