Question 765440

{{{f(x) = 4x^3 + 7x^2 + 6x - 2}}}..writ {{{7x^2}}} as {{{8x^2-x^2}}} and {{{6x}}} as {{{8x-2x}}}

{{{f(x) = 4x^3+8x^2+8x-x^2-2x-2}}}.....group

{{{f(x) = (4x^3-x^2)+(8x^2-2x)+(8x-2)}}}

{{{f(x) = x^2(4x-1)+2x(4x-1)+2(4x-1)}}}

{{{f(x) = (4x-1)(x^2+2x+2)}}}

solutions:

{{{(4x-1)(x^2+2x+2)=0}}}

if {{{4x-1=0}}} => {{{4x=1}}} => {{{x=1/4}}}

now use quadratic formula to find remaining two roots from {{{(x^2+2x+2)=0}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-2 +- sqrt( 2^2-4*1*2 ))/(2*1) }}}

{{{x = (-2 +- sqrt( 4-8 ))/2 }}}

{{{x = (-2 +- sqrt( -4 ))/2 }}}

{{{x = (-2 +- 2i)/2 }}}

{{{x = -2/2 +- 2i/2 }}}

{{{x = -1 +- i }}}

roots are:

{{{x = -1 + i }}}

and

{{{x = -1 - i }}}


so, you have one real and two complex roots


{{{ graph( 600,600, -10,10, -10, 10, 4x^3 + 7x^2 + 6x - 2) }}}