How do you factor $n^{2} + 5 n + 7$ $n^2+5n+7$ ?

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Answer 1 · 2015-10-01T21:56:41

$n = (\frac{- 5 + \sqrt{3} i}{2}), (\frac{- 5 - \sqrt{3} i}{2})$ $n=((-5+sqrt(3)i)/2),((-5-sqrt(3)i)/2)$

$n^{2} + 5 n + 7$ $n^2+5n+7$ is a quadratic equation $a x^{2} + b x + c$ $ax^2+bx+c$ , where $a = 1, b = 5, c = 7$ $a=1, b=5, c=7$ .

You can use the quadratic formula to factor this quadratic equation.

Quadratic Formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Substitute $n$ $n$ for $x$ $x$ .

$n = \frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 1 \cdot 7}}{2 \cdot 1} =$ $n=(-5+-sqrt(5^2-4*1*7))/(2*1)=$

Simplify.

$n = \frac{- 5 \pm \sqrt{25 - 28}}{2} =$ $n=(-5+-sqrt(25-28))/2=$

Simplify.

$n = \frac{- 5 \pm \sqrt{- 3}}{2} =$ $n=(-5+-sqrt(-3))/2=$

Simplify.

$n = \frac{- 5 \pm \sqrt{3} i}{2} =$ $n=(-5+-sqrt(3)i)/2=$

$n = (\frac{- 5 + \sqrt{3} i}{2}), (\frac{- 5 - \sqrt{3} i}{2})$ $n=((-5+sqrt(3)i)/2),((-5-sqrt(3)i)/2)$

How do you factor n^2+5n+7n2+5n+7n^2+5n+7?