# Factorize or simplify and solve the following equation completely : (2x^2-1)^2-(5x+2)^2=0

*print*Print*list*Cite

`(2x^2-1)^2- (5x+2)^2=0`

This equation can be factorized into a different format but it would certainly not simplify it. For example, by multiplying out the brackets and rearranging:

`(2x^2-1)^2 - (5x+2)^2 ` becomes

`4x^4-4x^2+1-25x^2-20x-4` which can be rearranged:

`(4x^4-25x^2)-(4x^2-1)-(20x+4)` (Note the changed symbols with the negatives).

This then simplifies to:

`(2x^2-5x)(2x^2+5x) - (2x-1)(2x+1) - 4(5x+1)` as they are differences of squares.

When solving this, substitute values such as x=1, x=2 to find the first possible solution and then use division to solve. Substituting x=3 into the **original** equation(which is the simplified version anyway) or any of the equations above renders:

`(2(3)^2 -1)^2 -(5(3)+2)^2 =0`

`therefore (18-1)^2 - (17)^2=0` reveals that x=3 is a solution which by division renders (x-3)(4x^3+12x^2+7x+1).

**A solution for this equation is (x-3)`(4x^3+12x^2+7x+1)` **

**and x=3 is a root / solution.**

`(2x^2-1)^2-(5x+2)^2=0`

Use the formula `a^2-b^2=(a-b)(a+b)` to get,

`{(2x^2-1)-(5x+2)}{ {(2x^2-1)+(5x+2)}=0 `

`rArr (2x^2-5x-3)(2x^2+5x+1)=0`

Use middle term factor for the first term

`rArr (2x^2-6x+x-3)(2x^2+5x+1)=0`

`rArr {2x(x-3)x+1(x-3)}(2x^2+5x+1)=0`

`rArr (2x+1)(x-3)(2x^2+5x+1)=0`

To solve, put each of the terms equal to zero.

Therefore,

`(2x+1)=0`

`x=-1/2,`

`(x-3)=0,`

`x=3,`

`(2x^2+5x+1)=0`

Use the quadratic formula to obtain

`x=(-5+-sqrt(25-8))/(2*2)`

`=1/4(-5+-sqrt17)`

So, the values of x are: `-1/2, 3, -5/4+sqrt17/4, -5/4-sqrt17/4`