Question 1089867


{{{(2x - 7)(3x + 5) + ax + 31 = 0}}}

if one root is {{{-4}}}, we have

{{{(2(-4) - 7)(3(-4) + 5) + a(-4) + 31 = 0}}}

{{{(-8 - 7)(-12 + 5) -4a + 31 = 0}}}

{{{(-15)(-7) -4a + 31 = 0}}}

{{{105 -4a + 31 = 0}}}

{{{ -4a + 136 = 0}}}

{{{ 4a = 136 }}}

{{{ a = 34 }}}


{{{(2x - 7)(3x + 5) + 34x + 31 = 0}}}

{{{6x^2 - 11x - 35 + 34x + 31 = 0}}}

{{{6x^2+ 23x   - 4  = 0}}}

{{{6x^2+ 24x-x   - 4  = 0}}}

{{{(6x^2+ 24x)-(x + 4 ) = 0}}}

{{{6x(x+ 4)-(x + 4 ) = 0}}}

{{{(6x-1)(x + 4 ) = 0}}}

roots:

if {{{(6x-1) = 0}}}->{{{x=1/6}}}
if {{{(x + 4 ) = 0}}}->{{{x=-4}}}(given)

so, the other root is {{{x=1/6}}}