In order to factor an integer, we need to repeatedly divide it by the ascending sequence of primes (2, 3, 5...). The number of times that each prime divides the original integer becomes its exponent in the final result. In our example, Prime number 2 to the power of 1 equals 2. Prime number 3 to the power of 1 equals 3. |
We need to factor the GCF (Greatest Common Factor). The resulting term is a product of the GCF and the original expression divided by the GCF. In our example, the GCF is equal to 3. |
We need to reduce this fraction to the lowest terms. This can be done by dividing out those factors that appear both in the numerator and in the denominator. In our example, this is the common factor: 3. |
We need to reduce this fraction to the lowest terms. This can be done by dividing out those factors that appear both in the numerator and in the denominator. In our example, this is the common factor: 3. |
Number 1 as a factor, does not need to be explicitly written. In other words: 1A=A. In our example, the above transformation has been applied once. |
Number 1 as a factor, does not need to be explicitly written. In other words: 1A=A. In our example, the above transformation has been applied once. |
Any fraction which has its denominator equal to 1 is equal to its numerator. In other words: NUM1=NUM In our example, the numerator is equal to x2. |
Any fraction which has its denominator equal to 1 is equal to its numerator. In other words: NUM1=NUM In our example, the numerator is equal to 2. |
In order to solve this non-linear inequality, we will apply the zero product theorem. That means that each variable factor in the original inequality needs to be set to zero. In our example, that will produce one new equation that can be solved. After this equation is solved, we can determine the solution intervals of the original inequality. |
Currently unable to solve inequalities of this type.