x2 + 100 has no real solutions. Applying the quadratic formula, we find two imaginary solutions: Zero plus or minus 10itimes the square root of 1.
x = 10i
x = -10i
where i is the square root of negative 1.
6(ab - ac + b2 - bc)
6ab(2x2+x-5)
(x - 5)(x - 3)
Remember to factor out the GCF of the coefficients if there is one. A perfect square binomial will always follow the pattern a squared plus or minus 2ab plus b squared. If it's plus 2ab, that factors to (a + b)(a + b) If it's minus 2ab, that factors to (a - b)(a - b)
(6x - 1)(6x - 1)
This expression factors as x -1 quantity squared.
2m^2 - 8 -First you should factor out a two. --> 2(m^2-4) -You now have something squared minus something else squared; You have m squared minus 2 squared. Whenever you have something squared minus something squared as you do in this case, there is a simple rule to remember: You can reduce that expression into the quantity of the square root of the first number or variable plus the square root of the second number or variable Times the quantity of the square root of the first number or variable minus the second number or variable squared. --> In the case of your expression: ----> 2(m+2)(m-2)<-----
There is no rational factorisation.
There are no rational factors.
(9 - y)(a + 5)
(a - 2)(a^2 + 6)
(b-c)(a+b)-ac
(3y - 5)(y + 5)
(x + 3)(3x - 2)
68.3
(7a - 1)(6a + 1)
it is (x-y)(x-y) :)