Answer:
The equation has 5 solutions and its real soluations are x=3, +sqrt2, -sqrt2, and -1
Step-by-step explanation:
Answer:
x = 3, -1, 2, -2
Step-by-step explanation:[tex]x^{2}[/tex]
6([tex]x[/tex]-3)([tex]x^{2}[/tex]+4)([tex]x[/tex]+1) = 0
6([tex]x[/tex]-3)([tex]x[/tex]+2)([tex]x[/tex]-2)([tex]x[/tex]+1) = 0
Using the fact that ([tex]x^{2}[/tex]+4) = ([tex]x[/tex]-2)([tex]x[/tex]+2) - known as the difference of two squares rule
Thus solve by inserting values of [tex]x[/tex] that will result in the equation equating to 0.
ie: if [tex]x[/tex] = 3,
6(3-3)(3+2)(3-2)(3+1) = 0
0=0
RHS = LHS therefore true
