How to solve this quadratic expression by factoring? n(n +2)

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How to solve this quadratic expression by factoring? n(n +2)

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Answer 1 · 2018-07-16T10:34:36

$(n + 1 - 1) (n + 1 + 1)$ $(n+1-1)(n+1+1)$

Explanation:

$n^{2} + 2 n + 1 - 1 = {(n + 1)}^{2} - 1^{2}$ $n^2+2n+1-1=(n+1)^2-1^2$

remember that

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$
we get

$(n + 1 - 1) (n + 1 + 1)$ $(n+1-1)(n+1+1)$
so $n = 0$ $n=0$ or $n = - 2$ $n=-2$

Answer 2 · 2018-07-16T10:40:04

$n = - 2 or n = 0$ $n=-2" or "n=0$

Explanation:

$the expression is given in factored form$ $"the expression is given in factored form"$

$to solve set equal to zero$ $"to solve set equal to zero"$

$n (n + 2) = 0$ $n(n+2)=0$

$equate each factor to zero and solve for n$ $"equate each factor to zero and solve for n"$

$n + 2 = 0 \Rightarrow n = - 2 or n = 0$ $n+2=0rArrn=-2" or "n=0$

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How to solve this quadratic expression by factoring? n(n +2)

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