Question

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

Answer 1

`(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)`

Answer 2

`(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]