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)#