Factor: 16x fourth - 40x square + 9?

Question

Show work on how answer was achieved.

Answer 1

X = you need to do your own homework lol

Answer 2

16 x^4 - 40x^2 + 9 = 0
x^2 (16 x^2-40) + 9 = 0
y (16y-40) + 9 =0

X^ 2 = y if ====> common points = 16/4 40 /4 9/4
4 10 2,25
y=2,25 ========> x=1.5

Answer 3

Let me teach you a simple trick to see everything easier.
Let y= x^2 (x^2 simply means x to the power of 2, or x square)

Therefore, the equation becomes:
16y^2 - 40y + 9
= (16y - 1) (y - 9)

Now I replace y with x ^ 2, I get
= (16 x^2 - 1) (x^2 - 9)

Enjoy!! :)

Answer 4

actually u hav to convert it intoa quadratic eqation assume x^2 = y then solve it ur equation will reduce to 16y^2- 40Y+ 9=0 now using the formula (-b + sqr(b^2-4ac))/2a calcul;ate the factors and den the vaues of y will come equate x^2 = y and get the answer

Answer 5

16x^4 - 40x +9
=(4x^2 - 1) (4x^2 - 9)

Test out answer using FOIL
(4x^2 * 4x^2) + (4x^2 * (-9)) + (4x^2 * (-1)) + ((-1) * (-9))
16x^4 - 36x^2 - 4x^2 + 9
16x^4 - 40x^2 + 9

Answer 6

16x^4 - 40x^2 + 9 =
(4x^2 - 2x - 6x + 3)(4x^2 + 6x + 2x + 3) =
(4x^2 - 8x + 3)(4x^2 + 8x + 3) =
(2x - 3)(2x - 1)(2x + 1)(2x + 3)

x = (3/2), (1/2), (-1/2), (-3/2)

Answer 7

Like what others told, let x^2=y:
A=4y^2-40y+9=(2y)^2-20(2y)+3^2
A=(2y)^2-2*3(2y)+3^2-14(2y)
A=(2y-3)^2-14(2y)
Now, substitute x^2=y:
A=(2x^2-3)^2-14(2x^2)=(2x^2-3)^2-(3xsqrt2)^2
A=(2x^2-3xsqrt2-3)(2x^2+3xsqrt2-3)