Question 1174668


{{{10x^2+tx+8 }}}

compare to {{{ax^2+bx+c}}}, =>{{{ a=10}}}, {{{b=t}}}, and {{{c=8}}}

you need

{{{ac=10*8=80}}}

factors of {{{80}}} are:±{{{1}}} | ±{{{2}}} | ±{{{4 }}}| ±{{{5}}} | ±{{{8 }}}| ±{{{10}}} | ±{{{16 }}}| ±{{{20}}} | ±{{{40}}} | ±{{{80 }}}

following sums are potential values for {{{t}}}

{{{1+80=81}}}
{{{2+40=42}}}
{{{4+20=24}}}
{{{5+16=21}}}
{{{8+10=18}}}

same is with negative values
 
so {{{t}}}= ±{{{18}}},±{{{21}}},±{{{24}}},±{{{42}}},±{{{81}}}

check some:

{{{t}}}= ±{{{18}}}

if {{{t= 18}} we have

{{{10x^2+18x+8}}} ....we can write {{{18x}}} as {{{10x +8x}}}
{{{10x^2+10x +8x+8}}}...group
{{{(10x^2+10x) +(8x+8)}}}
{{{10x(x+1) +8(x+1)}}}
{{{(10x +8)(x+1)}}}

if {{{t=-18}}}

{{{10x^2-18x+8{{{ ....we can write {{{-18x }}}as {{{-10x -8x}}}

{{{10x^2-10x -8x+8}}}...group
{{{(10x^2-10x) -(8x-8)}}}
{{{10x(x-1) -8(x-1)}}}
{{{(10x -8)(x-1)}}}

{{{t}}}= ±{{{21}}}

if {{{t=21 }}}which can be written as {{{16+5}}}

{{{10x^2+21x+ 8}}}
{{{10x^2+16x+5x+ 8}}}
{{{(10x^2+16x)+(5x+ 8)}}}
{{{2x(5x+8)+(5x+ 8)}}}
{{{(2x + 1) (5x + 8) }}}

 do same using {{{t=-21}}}, and you will get  {{{(2 x - 1) (5 x - 8) }}}