factor completely
-36c^4 + 289c^2 -400
-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)
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)
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
To factor the expression completely, we need to first recognize that it is a quadratic equation in terms of "c^2," which means that we can use the quadratic factoring technique.
Let's start by rewriting the equation:
-36c^4 + 289c^2 - 400
Now, let's factor the equation:
-36c^4 = (-6c^2)^2
289c^2 = (17c)^2
-400 = -20 * 20
So, our equation can be rewritten as:
(-6c^2)^2 + 2(17c)(-6c^2) + (17c)^2 - 20 * 20
Now, we can use the quadratic factoring technique to fully factor the expression. The goal is to find two binomials whose product equals the given quadratic.
Let's rewrite the expression, but this time considering it as a quadratic in terms of "c^2":
((-6c^2) + (17c))^2 - 20^2
Now, we can solve it as the difference of squares:
(((-6c^2) + (17c) + 20)((-6c^2) + (17c) - 20)
Now, let's factor each binomial separately:
((-2c - 5)(3c - 4))((-2c + 5)(3c + 4))
Therefore, the expression -36c^4 + 289c^2 - 400 can be factored completely as:
(-2c - 5)(3c - 4)(-2c + 5)(3c + 4)