Question

How do i factor n2 - n - 56

Answer 1

`{n}^(2) - n - 56 =`

`(n - 8)(n + 7)`

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`a {x}^(2) + bx + c = 0`

`delta = {b}^(2) - 4ac`

`x(1) = ( - b + √(delta) )/(2a) \\`

`x(2) = ( - b - √(delta) )/(2a) \\`

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`{n}^(2) - n - 56 = 0`

`delta = ({ - 1})^(2) - 4(1)( - 56)`

`delta = 1 + 224`

`delta = 225`

Thus ;

`n(1) = ( - ( -1) + √(225) )/(2 * 1) \\`

`n(1) = (1 + 15)/(2) \\`

`n(1) = (16)/(2) \\`

`n(1) = 8`

And

`n(2) = ( - ( - 1) - √(225) )/(2 * 1) \\`

`n(2) = ( 1 - 15)/(2) \\`

`n(2) = - (14)/(2) \\`

`n(2) = - 7`

So

`n(1) = 8`

`n(1) - 8 = 0 \: \: \: (\alpha )`

`n - 8 = 0 \: \: \: \: ( \alpha )`

And

`n(2) = - 7`

`n(2) + 7 = 0 \: \: \: \: \: ( \beta )`

`n + 7 = 0 \: \: \: \: ( \beta )`

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`\alpha * \beta =`

`(n - 8)(n + 7)`