How do you factor #3x^4-2x^2-1=0#?

2 Answers
Jim G.
May 4, 2018

#(3x^2+1)(x-1)(x+1)=0#

Explanation:

#"using the a-c method of factoring the quadratic in "x^4#

#"the factors of the product "3xx-1=-3#

#"which sum to - 2 are - 3 and + 1"#

#"split the middle term using these factors"#

#3x^4-3x^2+x^2-1larrcolor(blue)"factor by grouping"#

#=color(red)(3x^2)(x^2-1)color(red)(+1)(x^2-1)#

#"take out the "color(blue)"common factor "(x^2-1)#

#=(x^2-1)(color(red)(3x^2+1))#

#x^2-1" is a "color(blue)"difference of squares"#

#"and factors in general as"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#rArrx^2-1=(x-1)(x+1)#

#rArr3x^4-2x^2-1=(3x^2+1)(x-1)(x+1)#

Binayaka C.
May 4, 2018

#(x-1)(x+1)(3 x^2+1)=0 #

Explanation:

#3 x^4-2 x^2 -1 =0 # or

#3 x^4-3 x ^3 +3 x^3 -3 x^2 + x^2 -1 =0 # or

#3 x^3(x-1)+3 x ^2(x-1)+ (x+1)(x-1)=0 # or

#(x-1)(3 x^3+3 x ^2+ x+1)=0 # or

#(x-1){3 x^2(x+1)+1( x+1)}=0 # or

#(x-1)(x+1)(3 x^2+1)=0 # [Ans]

Related questions
See all questions in Factoring Completely
Impact of this question
3840 views around the world
You can reuse this answer
Creative Commons License