In one bracket, write 16 times the square root of 2 minus 3x. In the other one, write 16 times the square root of 2 plus 3x.
3(x2 + 3x + 1)
(2x+3)(x+3)
(x + 12)(x - 3)
Factor as : (x+9)(x+10) using the first terms ( x terms) multiply to x2; the last terms multiply to 90; the sum of 9x + 10 x = 19x
it is a combination of perfect squares the square of 9x is squared is +/- 3x, and the square of 4 is +/- 2 so the answer is (3x+2)(3x-2) which is 9xsquared-6x+6x-4 =9xsquared-4
Type your answer here... 81
9x2 + 8x - 1 = (9x - 1)(x + 1)
3(x2 + 3x + 1)
(x+1)(x+8)
(3x - y+ 2)(3x + y + 2)
You don't need to factor that; you can add them together, since they have the same variable.However, assuming you might mean 81x squared + 9x, you can take out the common factor, which is 9x.
x2 - 9x + 8 = (x-1)(x-8)
(9x)2 +9x/(45x)2 +81= 9+9x/86
(2x+3)(x+3)
x2-9x+8=0 has two solutions:x = 8x = 1
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)
9x squared plus 16 = 0 factored is plus and minus 4/3 i.