-36c^4 + 289c^2 -400
= -(36x^4 - 289x^2 + 400)
sure would be nice if that middle term was -240x^2, then we would have a perfect square.
Well, no big deal, let's make it that
= -(36x^4 - 289x^2 + 400 +49x^2 - 49x^2)
= 49x^2 - (36x^4 - 240x^2 + 400)
= (7x)^2 - (x^2 - 20)^2 , ahhh, a difference of squares
= (7x + x^2 - 20)(7x - x^2 + 20)
factor completely
-36c^4 + 289c^2 -400
3 answers
let x = c^2
-1 [ 36 x^2 - 289 x + 400 ]
I have to solve quadratic to factor
36 - 289 x+400 = 0
x = [ 289 +/- sqrt (83521-57600)]/72
= [289 +/- sqrt(25921)]/72
= [289 +/- 161 ]/72
= 128/72 OR 450/72
= 16/9 OR 25/4
SO
(x -16/9) and (x-25/4) are factors
so
(-1) (9x-16)(4x-25) and I divided by 8 and 18
so
-1 (9x-16)(4x-25)
but x = c^2
-1 (9c^2-16)(4c^2-25)
-1 (3c-4)(3c+4)(2c-5)(2c+5)
-1 [ 36 x^2 - 289 x + 400 ]
I have to solve quadratic to factor
36 - 289 x+400 = 0
x = [ 289 +/- sqrt (83521-57600)]/72
= [289 +/- sqrt(25921)]/72
= [289 +/- 161 ]/72
= 128/72 OR 450/72
= 16/9 OR 25/4
SO
(x -16/9) and (x-25/4) are factors
so
(-1) (9x-16)(4x-25) and I divided by 8 and 18
so
-1 (9x-16)(4x-25)
but x = c^2
-1 (9c^2-16)(4c^2-25)
-1 (3c-4)(3c+4)(2c-5)(2c+5)
Damon's answer is correct.
Just looked at mine and I forgot the 6 in front of the x^2 term
second last line should have been
= (7x)^2 - (6x^2 - 20)^2 , ahhh, a difference of squares
= (7x + 6x^2 - 20)(7x - 6x^2 + 20)
= (3x-4)(2x+5)(5-2x)(4+3x)
the same answer as Damon
Just looked at mine and I forgot the 6 in front of the x^2 term
second last line should have been
= (7x)^2 - (6x^2 - 20)^2 , ahhh, a difference of squares
= (7x + 6x^2 - 20)(7x - 6x^2 + 20)
= (3x-4)(2x+5)(5-2x)(4+3x)
the same answer as Damon
2 answers