Question

factor x^4-(4x-5)^2

Answer 1

its the difference of 2 squares

Answer 2

2?

Answer 3

a^2 - b^2 , how do u factor it out?

Answer 4

multiplying?

Answer 5

(x-1) (x+5) (x^2-4 x+5)

Answer 6

(x-1) (x+5) (x+1)(x-5)

Answer 7

x^4-(4x-5)^2
x^4-(16x^2-40x+25)
x^4-16x^2+40x-25

Answer 8

is this how you completely factor it? Nick Black

Answer 9

nope

Answer 10

No what nick did was expand it or multiply it out,
factoring makes it a product of terms

Answer 11

(x-1) (x+5) (x+1)(x-5)
this is completely factored

Answer 12

how did u get this dumbcow

Answer 13

how you get that dumbcow?:p

Answer 14

haha my bad, im wrong
victor was right
(x-1) (x+5) (x^2-4 x+5)

Answer 15

difference of squares
a^2-b^2 = (a-b)(a+b)

Answer 16

factor x^4-(4x-5)^2
Difference of two squares is such that a^2 -b^2 0= (a+b)(a-b)
Use this to factor the orginal expression...
(x^2 +[4x-5])(x^2 -[4x-5])
Expand inner brackets and simplify...
(x^2 +4x-5)(x^2 -4x+5)

These brackets can again be factotrised

Answer 17

x^4-16x^2+40x-25
=(x^2-5)(x^2+5)+8x(5-2x)

Answer 18

so that's the answer?

Answer 19

(x^2)^2 -(4x-5)^2 = [x^2-(4x-5)][x^2+(4x-5)]
= [x^2-4x+5][x^2+4x-5] = [x^2-4x+5][x^2+5x-x-5]
= [x^2-4x+5][x(x+5)-1(x+5)] = [x^2-4x+5][(x-1)(x+5)]
= [x^2-4x+5](x-1)(x+5)

[x^2-4x+5] can again be factorized in the form (x-a)(x-b) where
a, b are complex numbers. Do you want them also to be factored ?

Answer 20

Are you away ? This is the right answer. You will need factors
containing complex nos only if you have learnt them.

Answer 21

If ax^2+bx+c = 0 ;

x = [-b +or- sqrt (b^2- 4ac)] / 2a

Using this, if x^2-4x+5 =0 [ here a=1, b=-4, c=5]

= [-(-4) +or- sqrt {(-4)^2- 4x1x5)] / 2x1

= [4 +or- sqrt (16-20)] / 2 = [4 +or- sqrt (-4)] / 2

= [4 +or- 2sqrt (-1)] /2 = [2 +or- sqrt (-1)]

= 2 + i or 2 - i . So the complete factorization is

(x^2)^2 -(4x-5)^2 = [x^2-(4x-5)][x^2+(4x-5)]
= [x^2-4x+5][x^2+4x-5] = [x^2-4x+5][x^2+5x-x-5]
= [x^2-4x+5][x(x+5)-1(x+5)] = [x^2-4x+5][(x-1)(x+5)]
= [x^2-4x+5](x-1)(x+5)
= [ {x-(2+i)}{x-(2-i)}](x-1)(x+5)
= {x-2-i)}{x-2+i)(x-1)(x+5)

Answer 22

Note: i stands for sqrt(-1)

Note: sqrt(-4) = sqrt [4(-1)] = sqrt(4) x sqrt(-1)

= 2 sqrt(-1) = 2i

Answer 23

Contact me at acshyamraj@gmail.com

Answer 24

i dont think i've learned this:p
its for my pre-ap calculus summer packet and i just dont know how to completely fsctor it out