No, 18 is a multiple of 3. 3 is a factor of 18.
X^2 + 7X + 18 = 0 try the discriminant b^2 - 4ac 7^2 - 4(1)(18) 49 - 72 < 0 so, no real roots and no factors of 18 add to 7
No, 18 is a multiple of 9. (18 = 2 x 9, so 2 and 9 are factors of 18).
16 + 16 + 16 + 17 + 18 + 18 + 18 + 19 + 19 + 20 + 20 + 20 + 21 = 238
If you're solving for b, then you use the quadratic formula to factor out b. In this case, b= -1/4 or -18/12
6(w + 3)
3
I'm not sure this was notated correctly. As written, this is -11x plus or minus 18, which doesn't factor.
No, a factor cannot be larger than the numbers it is compared to.
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)
(x^18 - 6x^9 + 18)(x^18 + 6x^9 + 18)
As written, that's 12a + 18 which factors to 6(2a + 3) a^2 + 11a + 18 = (a + 9)(a + 2)
-((x + 2)(x - 9))
You take out the common factor, 6. 6y + 18 = 6(y + 3)
The GCF is 6.
4x + 27x + 18 = 0 31x + 18= 0 31x + 18 - 18= 0 - 18 31x = -18 31x / 31 = -18 / 31 x = -.58
5x squared plus 33x plus 18 = (5x + 3)(x + 6) x = -6, -3/5