A question?

Question

Factorise this: 4-8x+20x^2

my tutor says this isn't possible.. I factorised it and my tutor says that my answer is wrong, even though i checked it 15 times....

Try it and see if it is possible to factorise it..

Answer 1

20x² - 8x + 4 = 0 Divide both sides by 4.
5x² - 2x + 1 = 0

To see if it's factorable, look at the "discriminant" of the quadratic equation:

b² - 4ac = (-2)² - 4(5)(1) = 4 - 20 = -16

Since the discriminant is negative (that's the part under the square root in the quadratic equation), you can say:

-- There is no solution (over the real numbers).
-- There are two complex conjugate solutions (over the complex numbers).

If you don't know what complex numbers are, then go with the first answer.

Answer 2

It factors into 4(5x^2 - 2x + 1), but some would call this a trivial factorisation.

5x^2 - 2x + 1 does not have real factors, only the imaginary factorisation (5x - 1 - 2i)(5x - 1 + 2i)/5, because its "discriminant" is 2^2 - 4*5*1 which is negative. For the factors to be real, the discriminant needs to be positive; and for them to be rational, it needs to be a perfect square.

Answer 3

first you can factorize by 4

(1-2x +5x^2) *4 And if you see tere are no roots to that equation since the discriminant 4 -20 = -16 is negative.

You can perhaps factorize in a complex number

Answer 4

that is factorable scientifically.

arranged in ascending order.

20x^2 - 8x + 4

get the common factor 4 w/c is the LCM

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


so that's it! I'm sure about it!

Answer 5

4-8x+20x^2 = 20x^2 - 8x + 4 =0 this has complex conjugate roots
x1=(1/5)+(2/5)*i and x2 = (1/5)-(2/5)*i

This is because for equation ax^2 + bx + c= 0
The roots are x = [-b (+or-) sqrt(b^2 - 4ac)] / 2a

Comparing with 20x^2 - 8x + 4=5x^2 - 2x + 1=0, we get
a = 5, b=-, c=4

So, x = [-(-8) (+or-) sqrt((-8)^2 - 4*20*4)] / 2*20
=> x = [8 (+or-) sqrt(64 - 320)] / 40
=> x = [8 (+or-) 16sqrt(-1)] / 40
=> x = 8[1 (+or-) 2sqrt(-1)] / 40
=> x = [1 (+or-) 2sqrt(-1)] / 5

So, the factors are (x - {(1/5)+(2/5)*i}) and (x - {(1/5)-(2/5)*i})

I hope, I answered your question.

All the best

Answer 6

4 - 8x + 20x²

- - - - - - -

Rearrange the equation to:

20x² - 8x + 4

4(5² -2x + 1)

The common factor is 4

- - - - - - - -s-

Answer 7

yes, it's impossible to factorise because if you use the quadratic equation, the square root is negative, meaning there's no solution.

Answer 8

yup, i think its impossible to factorized...
rewriting your equation,

20x^2-8x+4 = 0, i only equate to zero for ease.
dividing by 5 you'll get,
4x^2-2x+1 = 0... way too impossible...

Answer 9

A complex number

x=1/5-2i/5 and x=1/5+2i/5

Answer 10

This is surely not factorisible . OK tell me your answer