We need to get rid of parentheses in this term. All the negative factors will change sign. In our example, we have only one negative factor. The sign of the term will change, since there is an odd number of negative factors. |
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 7 to the power of 1 equals 7. |
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 3 equals 8. |
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 5 equals 32. |
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 2 equals 4. Prime number 3 to the power of 1 equals 3. |
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 2 equals 9. |
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 2. |
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 2·3x. |
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: 2. |
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: 2. |
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: 2. |
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: 2·3x. |
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: 2·3x. |
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. |
Numerical terms in this expression have been added. |
Numerical terms in this expression have been added. |
Numerical terms in this expression have been added. |
Numerical terms in this expression have been added. |
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 twice. |
Numerical terms in this expression have been added. |
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 twice. |
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 twice. |
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 7x6. |
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 24·x3. |
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 22. |
This is a special case of exponentiation. The following rule is applied: A1=A In our example, A is equal to 2. |
This is a special case of exponentiation. The following rule is applied: A1=A In our example, A is equal to x. |
This is a special case of exponentiation. The following rule is applied: A1=A In our example, A is equal to 3. |
We need to evaluate a power by multiplying the base by itself as many times as the exponent indicates. In our example, base 2 will be multiplied by itself four times. |
We need to evaluate a power by multiplying the base by itself as many times as the exponent indicates. In our example, base 2 will be multiplied by itself twice. |
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 2x. |
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 3. |
We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, the following 2 terms will change sign: 16, x3 |
We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, the following 2 terms will change sign: 2, x |
We can reduce this fraction by dividing both numerator and denominator by a common numeric factors. In our example, both number 2 in numerator and number 4 in denominator are divisible by 2. |
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: 2x. |
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 twice. |
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 twice. |
In order to factor a quadratic trinomial of the following form: AX2+BX+C, we first have to make the square in the first term explicit. In our example, X is equal to x3. |
We need to factor this trinomial by grouping. The first step in this process consists of splitting the middle term into two terms. In our example, −16(x3) has been split into −14x3 and −2x3. Notice that one common factor can be factored from the first 2 terms, and another from the second 2 terms |
In order to factor this polynomial, we need to independently factor the first two and the last two terms. This process will create two new terms containing a common factor. |
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: 7x3. |
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: 7x3. |
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: 2. |
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: 2. |
Numerical terms in this expression have been added. |
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. |
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 twice. |
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. |
Numerical terms in this expression have been added. |
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 x3. |
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. |
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 x3. |
This is a special case of exponentiation. The following rule is applied: A1=A In our example, A is equal to 2. |
We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, the following 2 terms will change sign: x, 3 |
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. |
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 x3−2. |
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: x3−2. |
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: −x3+2. |
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 7x3. |
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. |
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: x3−2. |
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. |
Numerical terms in this expression have been added. |
We need to perform a multiplication. The following rule is applied: ABC=ACB In our example, the factors in the new numerator are: (4x3+9x2+4), (7x3−2), Notice that all non-fraction factors are placed in the numerator. The factors in the new denominator are: 2, x2, (x3−2)3, |
We need to add fractions. The following rule is applied: AB+CD=LCDBA+LCDDCLCD This example involves 2 terms. The LCD is equal to: (x3−2)32x2 |
Numerical terms are commonly written first. |
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. |
Numerical terms are commonly written first. |
We need to get rid of parentheses in this term. All the negative factors will change sign. In our example, we do not have any negative factors. The sign of the term will not change. |
We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
Numerical terms are commonly written first. |
Numerical factors in this term have been multiplied. |
We need to arrange this term for expansion. Commonly, the non-expandable factor is written first. In our example, the factor is 6x2. |
We need to expand this term by multiplying two expressions. The following product distributive property will be used: (A+B)(C+D)=AC+AD+BC+BD. In our example, the resulting expression will consist of 6 terms: the first term is a product of 4x3 and 7x3. the second term is a product of 4x3 and −2. the third term is a product of 9x2 and 7x3. the fourth term is a product of 9x2 and −2. the fifth term is a product of 4 and 7x3. the sixth term is a product of 4 and −2. |
We need to organize this term into groups of like factors, so we can combine them easier. Numerical terms are commonly written first. The following are like factors: x3, x3 |
Numerical terms are commonly written first. |
We need to organize this term into groups of like factors, so we can combine them easier. Numerical terms are commonly written first. The following are like factors: x2, x3 |
Numerical terms are commonly written first. |
Numerical factors in this term have been multiplied. |
Numerical factors in this term have been multiplied. |
We need to combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1. Numerical factors will be multiplied. The following are like factors: x3, x3 |
Numerical factors in this term have been multiplied. |
We need to combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1. Numerical factors will be multiplied. The following are like factors: x2, x3 |
Numerical factors in this term have been multiplied. |
Numerical terms in this expression have been added. |
Numerical terms in this expression have been added. |
We need to organize this expression into groups of like terms, so we can combine them easier. There is only one group of like terms: −8x3, 28x3 |
We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There is only one group of like terms: −8x3, 28x3 |
We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
We need to start expanding this term by multiplying its first two factors (a term and an expression). The following product distributive property will be used: A(B+C)=AB+AC. In our example, the resulting expression will consist of 2 terms: the first term is a product of 6x2 and x3. the second term is a product of 6x2 and −2. |
We need to combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1. The following are like factors: x2, x3 |
We need to get rid of parentheses in this term. All the negative factors will change sign. In our example, we have only one negative factor. The sign of the term will change, since there is an odd number of negative factors. |
Numerical terms in this expression have been added. |
Numerical terms are commonly written first. |
Numerical factors in this term have been multiplied. |
We need to expand this term by multiplying two expressions. The following product distributive property will be used: (A+B)(C+D)=AC+AD+BC+BD. In our example, the resulting expression will consist of 4 terms: the first term is a product of 6x5 and −2x. the second term is a product of 6x5 and −3. the third term is a product of −12x2 and −2x. the fourth term is a product of −12x2 and −3. |
We need to organize this term into groups of like factors, so we can combine them easier. Numerical terms are commonly written first. The following are like factors: x5, x |
Numerical terms are commonly written first. |
We need to organize this term into groups of like factors, so we can combine them easier. Numerical terms are commonly written first. The following are like factors: x2, x |
Numerical terms are commonly written first. |
We need to combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1. Numerical factors will be multiplied. The following are like factors: x5, x |
Numerical factors in this term have been multiplied. |
We need to combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1. Numerical factors will be multiplied. The following are like factors: x2, x |
Numerical factors in this term have been multiplied. |
Numerical terms in this expression have been added. |
Numerical terms in this expression have been added. |
We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
We need to organize this expression into groups of like terms, so we can combine them easier. There are 4 groups of like terms: first group: −12x6, 28x6 second group: −18x5, 63x5 third group: 24x3, 20x3 fourth group: 36x2, −18x2 |
We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There are 4 groups of like terms: first group: −12x6, 28x6 second group: −18x5, 63x5 third group: 24x3, 20x3 fourth group: 36x2, −18x2 |
Since the right side of this equation is equal to zero, we can get rid of the denominator on the left side. |