3xx-588
3(xx-196)
Factor (long):
xx+0x-196
axx+bx+c
a=1
b=0
c=-196
Multiply a and c:
1*-196=-196
Find multiples of ac that add up to b (d,e [ie de=ac, d+e=b]):
-196=-1*2*2*7*7,
-2*98, 2*-98
-7*28, 7*-28
-4*49, 4*-49
-14*14,
Rewrite axx+bx+c into axx+dx+ex+c:
3(xx-14x+14x-196)
Factor sets of terms (ie axx+dx, and ex+c):
3(xx-14x+14x-196)
3(x(x-14)+14(x-14))
Combine like terms:
3(x(x-14)+14(x-14))
3(x-14)(x+14)
Factor (short):
3(xx+0x-196)
Factors of -196, find which ones add up to 0
-196=-1*2*2*7*7,
-2*98, 2*-98
-7*28, 7*-28
-4*49, 4*-49
-14*14,
Rewrite in form of (x+d)(x+e)
3(x+14)(x-14)
Factor (shorter):
-when a polynomial takes the form of xx-c, it can be factored into the form:
(x+sqrt(c))(x-sqrt(c))
sqrt(196)=14,-14
3(x+14)(x-14)
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (0 plus or minus the square root of 30) divided by 3.
x = 1.8257418583505538
x = -1.8257418583505538
2(5r + 6)(5r - 6)
(2x - 3)(3x + 2)
6(ab - ac + b2 - bc)
3X2 - 123(X2 - 4)=========All the factoring that can be done here.
if by that you mean 3x - 125 well you can't. unless you want to get a fraction or a decimal which is not the whole point of factorizing you could facorising 3x -18 by dividing each of the number by 3 eg. 3x-18 =3(x-6)
(x + 3)(3x - 2)
(x - 6)(x + 9)
Yes, but not in rational terms. The factorisation is: {x - [3-sqrt(57)]/6} * {x - [3+sqrt(57)]/6}
3x2 -14x -24 can be factored as (3x + 4)(x - 6).
That factors to (x - 3)(x + 6)
Your factors must begin 3x and x to give 3x2. The factors of 12 are 1/12, 2/6 and 3/4. Which pair, together with 3x and x, can add to -5? 3 and 4 seem to fit the bill: (3x + 4)(x - 3)
(x - 2)(2x + 3)
(x + 6)(x - 6)
2(5r + 6)(5r - 6)
(5x - 6)(5x + 6)
y² - y - 6 = (y + 2)(y - 3)
(2x - 3)(3x + 2)