Find all integers $n$ such that the quadratic $7x^2 + nx - 11$ can be expressed as the product of two linear factors with integer coefficients.

The all integer values of n are - 76, - 4, 4, 76

Step-by-step explanation:

∵ The factors of 7 are 1 and 7

∵ The factors of 11 are 1 and 11

- The last term of the quadratic is negative, that means the two

linear factors have different middle sign

∴ The factors are (7x ± 1) (x ± 11) OR (7x ± 11) (x ± 1)

Let us take one by one

The middle term of the quadratic is the sum of the product of nears and ext-reams

In (7x + 1) (x - 11)

∵ The product of ext-reams is 7x * - 11 = - 77x

∵ The product of nears is 1 * x = x

∵ Their sum is - 77x + x = - 76x

∴ n = - 76

In (7x - 1) (x + 11)

∵ The product of ext-reams is 7x * 11 = 77x

∵ The product of nears is - 1 * x = - x

∵ Their sum is 77x + - x = 76x

∴ n = 76

In (7x + 11) (x - 1)

∵ The product of ext-reams is 7x * - 1 = - 7x

∵ The product of nears is 11 * x = 11x

∵ Their sum is - 7x + 11x = 4x

∴ n = 4

In (7x - 11) (x + 1)

∵ The product of ext-reams is 7x * 1 = 7x

∵ The product of nears is - 11 * x = - 11x

∵ Their sum is 7x + - 11x = - 4x

∴ n = - 4

The all integer values of n are - 76, - 4, 4, 76