The Prime Factors of the Binomials?

Question

what of the factors of these questions

1. 12x(2nd power)y(2nd power) - 27
2. (x{2nd power}-5x+8)(x{2nd power}+1)

THanks for your help

Answer 1

12x^2y^2-27
=3(4x^2y^2-9)
=3(2xy-3)(2xy+3)

(x^2-5x+8)(x^2+1)
=x^4+x^2-5x^3-5x+8x^2+8
=x^4-5x^3+9x^2+8
i dont think there is a prime factor for the second one...
it is a prime factor itself...

Answer 2

In the first case, once you divide the equation by 3, you get the difference of two squares.
Let us call them a^2 and b^2
(a^2 - b^2) = (a+b)(a-b)

Replace a^2 and b^2 with the actual squares.

In the second case, the first factor (x^2 -5x +8) is a quadratic with a simple solution:

ax^2 + bx + c = (x + (b+sqrt(b^2-4ac))/2a)(x - (b+sqrt(b^2-4ac))/2a)

Of course, if (b^2 - 4ac) is negative, then the quadratic is already prime.

(x^2 + 1) is irreducible in real numbers (in other words, it is a prime factor).

In complex numbers, where i = sqrt(-1), it can be rewritten:
(x^2 + 1) = (x^2 - (-1)) making it a difference of squares.

(x^2 - (-1)) = (x+i)(x-i).

If you can't use complex, then another option is to multiply out the expression,
(x^2 -5x +8)(x^2+1) = x^4 -5x^3 +9x^2 -5x +8
Then try to factor it.

Answer 3

The first one is common factoring followed by difference of squares:

3(4x^2 - 9)
=3(2x + 3)(2x - 3)

The second one does not factor. The first expression has a discriminant of 25 - 4(1)(8) which is negative. This means it will have complex roots only. The second expression is the sum of squares which is not factorable in the real numbers either. It too has complex roots only.