How do you factor $n^{2} - 5 n + 6$ $n^2-5n+6$ ?

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How do you factor $n^{2} - 5 n + 6$ $n^2-5n+6$ ?

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Answer 1 · 2015-05-16T10:57:15

Notice that $5 = 2 + 3$ $5 = 2+3$ and $6 = 2 \cdot 3$ $6 = 2*3$ , so

$(n - 2) (n - 3) = (n \cdot n) - (2 \cdot n) - (3 \cdot n) + (2 \cdot 3)$ $(n - 2)(n - 3) = (n*n)-(2*n)-(3*n)+(2*3)$

$= n^{2} - (2 + 3) n + (2 \cdot 3)$ $= n^2-(2+3)n+(2*3)$

$= n^{2} - 5 n + 6$ $= n^2-5n+6$

In general $(x + a) (x + b) = x^{2} + (a + b) x + a b$ $(x + a)(x + b) = x^2 + (a+b)x + ab$ so if you can spot two numbers $a$ $a$ and $b$ $b$ such that their sum is the coefficient of $x$ $x$ and their product is the constant term then you can quickly factorise such a quadratic formula.

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How do you factor $n^{2} - 5 n + 6$ $n^2-5n+6$ ?

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