Question 199937: 18x4-9x3+6x2 factor completely x is exponets
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! 18x^4 -9x^3 +6x^2
.
(3x^2)(6x^2 -3x +2)
.
3x^2 =0,,,,,,,x=0
.
using the quadratic formula,,,for 2nd factor
.
a=6,,,,b=-3,,,,c=2
.
x= [ -(-3) +/- { (-3)^2 - 4 (6) (2) }^.5 ] / 2 *6
.
x=[ 3 +/- { 9-48}^.5 ] /12
.
x= [ 3 +/- { -39}^.5] /12
.
x= 3/12 +/- (6.24/12)i
.
x = .25 +/- .5204i
.
x= .25 + .5204i,,,,,or x = .25 - .5204i,,,,,,also remember previous factor,,,3x^2
.
Factors are,,,3x^2,,,, (x-(.25+.5204 i ) ),,,( x-(.25 -.5204 i) )
.
3x^2,,,, x-.25 -.5204i,,,, x-.25+.5204 i
|
|
|
