Question

Сократи дробь 2x^2−12x+16/x^3−64

Answer 1

$\frac{2x^2-12x+16}{x^3-64}=\frac{2\, (x^2-6x+8)}{(x-4)(x^2+4x+16)}=\frac{2\, (x-2)(x-4)}{(x-4)(x^2+4x+16)}=\frac{2\, (x-2)}{x^2+4x+16}$

Answer 2

To simplify the fraction (2x^2 - 12x + 16) / (x^3 - 64), we first factor both the numerator and denominator:

Numerator (2x^2 - 12x + 16): Factor out the common factor 2: 2(x^2 - 6x + 8)

Now, we need to factor the quadratic expression inside the parentheses: 2(x^2 - 6x + 8) = 2(x - 4)(x - 2)

Denominator (x^3 - 64): This is a difference of cubes, which can be factored as follows: x^3 - 64 = (x - 4)(x^2 + 4x + 16)

Now, the original fraction becomes: (2x^2 - 12x + 16) / (x^3 - 64) = 2(x - 4)(x - 2) / ((x - 4)(x^2 + 4x + 16))

Next, we can cancel out the common factor (x - 4) from both the numerator and the denominator:

(2x^2 - 12x + 16) / (x^3 - 64) = 2(x - 2) / (x^2 + 4x + 16)

So, the simplified form of the fraction is 2(x - 2) / (x^2 + 4x + 16).