How do you factor $5x^2 + 9x - 81$ ?

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Answer 1 · 2015-05-31T22:52:32

$5x^2+9x-81$ is of the form $ax^2+bx+c$
with $a=5$ , $b=9$ , $c=-81$

The discriminant $Delta$ is given by the formula:

$Delta = b^2-4ac = 9^2-(4xx5xx-81) = 9^2 * 21$

$Delta > 0$ so $5x^2+9x-81 = 0$ has two distinct real roots. Unfortunately $Delta$ is not a perfect square, so those roots are irrational.

The roots of $5x^2+9x-81 = 0$ are given by the formula:

$x = (-b+-sqrt(Delta))/(2a)$

$= (-9+-9sqrt(21))/10$

$= 9/10(-1+-sqrt(21))$

Hence

$5x^2+9x-81$

$= 5(x - 9/10(-1+sqrt(21)))(x - 9/10(-1-sqrt(21)))$

$= 5(x + 9/10(1-sqrt(21)))(x + 9/10(1+sqrt(21)))$

How do you factor 5x^2 + 9x - 815x^2 + 9x - 81?