i hope you mean (2x^2 + 21x + 49) you cant solve it its already in its lowest possible form
(2x+7)(x+7)
3x+21x+9x=33x. Break it apart. The factor of x is x and the factors of 33 are 3 and 11. Therefore, the factors of 3x+21x+9x are 3, 11, and x.
3(7x + 11)
The answer will be 21x + 1
2x2+21x+49 = (2x+7)(x+7)
i hope you mean (2x^2 + 21x + 49) you cant solve it its already in its lowest possible form
2x2 + 21x + 49 = 2x2 +14x +7x + 49 = 2x(x + 7) + 7(x + 7) = (2x + 7)(x + 7)
2x2 + 21x + 49= 2x2 + 14x + 7x + 49= 2x(x + 7) + 7(x + 7)= (x + 7)(2x + 7)
2x2 + 21x + 49 = (2x + 7)(x + 7)
(2x+7)(x+7)
21x + 9 = 180 21x = 180 - 9 21x = 171 x = 171/21 x = 8.143
14xy - 21x + 7 '7' is a common factor to all three coefficients. So 'take it out'. Hence 7(2xy - 3x + 1) The rest of it , inside the brackets does not factor. However, I suspect the 'y' should be a 'power of 2' Hence 7(2x^(2) - 3x + 1) This factors to 7(2x - 1)(x - 1) Fully factored.!!!!
21x+452 = 32-84x 21x+84x = 32-452 105x = -420 x = -4
20x + 5 - 20 = 21x + 4 -x = 4 + 15 x = -19
3x+21x+9x=33x. Break it apart. The factor of x is x and the factors of 33 are 3 and 11. Therefore, the factors of 3x+21x+9x are 3, 11, and x.
234 it is not the answer