x^4 - x^2 plus 2 has no real rational roots.
x^4 - x^2 minus 2 factors to (x^2 + 1)(x^2 - 2)
4x2 + 5x + 1 - 4x2 - 6x = - x + 1 or 1 - x4x2 + 5x + 1 - 4x2 - 6x = - x + 1 or 1 - x4x2 + 5x + 1 - 4x2 - 6x = - x + 1 or 1 - x4x2 + 5x + 1 - 4x2 - 6x = - x + 1 or 1 - x
It can be. x^2 + x + 1 is a factor of 2x^2 + 2x + 2
The only factor is 2. 2*(t3 + 2t2 + 4x)
No
Since the problem has 4 terms, first you factor x cubed plus 9x squared, then you factor 2x plus 18. So when you factor the first two term, you would get x sqaured (x plus 9). Then when you factor the last two terms and you get 2 (x plus 9). Ypure final answer would be (x squared plus 2)(x plus 9)
2(3x^2 + 6x + 2)
factor the trinomial 16x^2+24x+9
(9a + 2)(9a + 2)
(2x + 1)(x + 2)
(r + 2)(r + 2)
(5x + 2)(x + 1)
The common factor is 2.