How do you factor $10 n - {(2 n - 3)}^{2}$ $10n-(2n-3)^2$ ?

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How do you factor $10 n - {(2 n - 3)}^{2}$ $10n-(2n-3)^2$ ?

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Answer 1 · 2018-03-11T01:03:56

First expand.
$10 n - 4 n^{2} - 6 n + 9$ $10n - 4n^2 -6n + 9$
$= - 4 n^{2} + 4 n + 9$ $= -4n^2 + 4n + 9$
Factorise out the -1.
$= - 1 (4 n^{2} - 4 n - 9)$ $= -1(4n^2 -4n -9)$
Now use quadratic equation.
a = 4, b = -4, c = -9.
$\frac{4 \pm \sqrt{16 + 144}}{- 72}$ $(4 +- sqrt(16+144))/-72$
Then you are left with (when simplified):
$\frac{1 \pm \sqrt{10}}{18}$ $(1 +- sqrt(10))/18$
Now put into factorised form...
$(18 n + 1 - \sqrt{10}) (n + 1 + \sqrt{10})$ $(18n +1 - sqrt(10))(n + 1 + sqrt(10))$

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How do you factor $10 n - {(2 n - 3)}^{2}$ $10n-(2n-3)^2$ ?

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