How do you factor the trinomial $2a^2+19a-10$ ?

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Answer 1 · 2016-07-14T10:36:23

$=(a+10)(2a-1)$

you can factor a trynomial

$ax^2+bx+c$

by applying the rule:

$ax^2+bx+c=a(x-x_1)(x-x_2)$

where $x_1 and x_2$ are the zeroes of the trynomial

So, you first solve the equation:

$2a^2+19a-10=0$

$a=(-19+-sqrt(19^2-4*2*(-10)))/(2*2)$

$a=(-19+-sqrt(361+80))/4$

$a=(-19+-sqrt(441))/4$

$a=(-19+-21)/4$

$a_1=-10 and a_2=1/2$

Then

$2a^2+19a-10=0=2(a+10)(a-1/2)$

$=(a+10)(2a-1)$

How do you factor the trinomial 2a^2+19a-102a^2+19a-10?