3x2 + 36x + 81 = 3(x2 + 13x + 27)
3x2 + 36x + 81 = 3(x2 + 12x + 27) = 3(x + 3)(x + 9), which are its prime factors; or, if you prefer, 3x2 + 36x + 81 = (3x + 9)(x + 9), which is also accurate. You may easily verify these results by multiplying back.
x2 + 10x + 21 = (x + 3)(x + 7) You need to find 2 numbers which sum to 10 and whose product of 21. In this case 3 + 7 = 10 and 3 x 7 = 21.
3x2 + 48 + 192 = 3x2 + 240 = 3 (x2+ 80)
3x2 + 10x - 8The auxiliary equation is 3x2 + 10x - 24 [because 8*3 = 24] Two factors of -24 which sum to 10 are 12 and -2. So the factorisation of the auxillary is (3x + 12)*(3x - 2) = 3*(x + 4)*(3x - 2) Dividing by the 3 used to create the auxiliary gives the solution as (x + 4)*(3x - 2).
-3x2+10x-3 -(3x2-10x+3) -(3x-1)(x-3)
3x2 + 10x + 3 = (x + 3)(3x + 1).
3x2 + 36x + 81 = 3(x2 + 13x + 27)
3x2 + 33x + 54 =3(x2 + 11x + 18) = 3(x + 9)(x + 2)
X + 4 is a factor of x^3 + 3x^2 - 10x - 24 along with (x + 2) and (x - 3)
3x2 + 36x + 81 = 3(x2 + 12x + 27) = 3(x + 3)(x + 9), which are its prime factors; or, if you prefer, 3x2 + 36x + 81 = (3x + 9)(x + 9), which is also accurate. You may easily verify these results by multiplying back.
5x + 10x doesn't need factorization as you can just add them up (5x+10x=15x). However, if you really want to, 5x+10x=5x(1+2)=5x(3)=15x
x2 + 10x + 21 = (x + 3)(x + 7) You need to find 2 numbers which sum to 10 and whose product of 21. In this case 3 + 7 = 10 and 3 x 7 = 21.
-8x + 3x2 - 3
6 ^ 3 2 answer:3x2
6 + 10x + 3 = 9 + 10x
3x2 + 48 + 192 = 3x2 + 240 = 3 (x2+ 80)