Up till now you can't but hopefully will be able to do so soon.
The answer depends on how good you are.You can either factorise the number completely and express it in terms of the exponents of its prime factors. Any prime with an exponent of 3 (or a multiple of 3) is a cube factor.A faster, but more demanding way is to see if 8, 27, 64 ... are factors.The answer depends on how good you are.You can either factorise the number completely and express it in terms of the exponents of its prime factors. Any prime with an exponent of 3 (or a multiple of 3) is a cube factor.A faster, but more demanding way is to see if 8, 27, 64 ... are factors.The answer depends on how good you are.You can either factorise the number completely and express it in terms of the exponents of its prime factors. Any prime with an exponent of 3 (or a multiple of 3) is a cube factor.A faster, but more demanding way is to see if 8, 27, 64 ... are factors.The answer depends on how good you are.You can either factorise the number completely and express it in terms of the exponents of its prime factors. Any prime with an exponent of 3 (or a multiple of 3) is a cube factor.A faster, but more demanding way is to see if 8, 27, 64 ... are factors.
Factor out each prime by prime to obtain: 4 x 5 = 2 x 2 x 5 So the answer is 2² x 5 * * * * * and the word is "superscript", not subscript.
HCF(735, 756) = 21 It can save you a lot of time if you use the fact that for any integers A and B, HCF(A, B) = HCF(A, B-A) That soon gives you small numbers to factorise.
If you know one linear factor, then divide the polynomial by that factor. The quotient will then be a polynomial whose order (or degree) is one fewer than that of the one that you stared with. The smaller order may make it easier to factorise.
Factorise fully is when brackets are involved in the equation
a²-a = a(a-1)
To factorise is to find the numbers that divide into the original number by only using prime numbers. For example factorise 20 = 2 times 2 times 5
to put into brackets
you do (245x)
The answer will depend on where the brackets are. In general the solution would be to expand all the brackets, combine like terms and then factorise.
6(t2s)
m(g-a)
-5
It is the opposite of Expanding The Brackets
qwertyuiopasdfjk
In the same way that you would factorise any other expressions that do not contain an equality sign depending on the expressions of which none have been given.