The way to simplify a rational expression; 1/4*z^2-1/6*z-(-5/24*z^2+1/12*z+(-23)/24) Step-by-Step Math Problem Solver

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Created on: 2011-12-12

Step 1

14z²−16z−(−524z²+112z−2324)

We need to perform a multiplication.

The following rule is applied:

ABC=ACB

In our example, the factors in the new numerator are:

1, z²,

Notice that all non-fraction factors are placed in the numerator.

The only factor in the new denominator is:

We need to perform a multiplication.

The following rule is applied:

ABC=ACB

In our example, the factors in the new numerator are:

1, z,

Notice that all non-fraction factors are placed in the numerator.

The only factor in the new denominator is:

We need to perform a multiplication.

The following rule is applied:

ABC=ACB

In our example, the factors in the new numerator are:

5, z²,

Notice that all non-fraction factors are placed in the numerator.

The only factor in the new denominator is:

We need to perform a multiplication.

The following rule is applied:

ABC=ACB

In our example, the factors in the new numerator are:

1, z,

Notice that all non-fraction factors are placed in the numerator.

The only factor in the new denominator is:

1z²4−1z6−(−5z²24+1z12−2324)

Step 2

1z²4−1z6−(−5z²24+1z12−2324)

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 once.

z²4−z6−(−5z²24+z12−2324)

Step 3

z²4−z6−(−5z²24+z12−2324)

We need to add fractions.

The following rule is applied:

AB+CD=LCDBA+LCDDCLCD

This example involves 3 terms.

The LCD is equal to:

24·12

z²4−z6−(12(−5z²)+24z+12(−23)24·12)

Step 4

z²4−z6−(12(−5z²)+24z+12(−23)24·12)

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.

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.

z²4−z6−(−12·5z²+24z−12·2324·12)

Step 5

z²4−z6−(−12·5z²+24z−12·2324·12)

Numerical factors in this term have been multiplied.

z²4−z6−(−60z²+24z−27624·12)

Step 6

z²4−z6−(−60z²+24z−27624·12)

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.

Prime number 5 to the power of 1 equals 5.

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.

Prime number 23 to the power of 1 equals 23.

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.

Prime number 3 to the power of 1 equals 3.

z²4−z6−(−2²·3·5z²+(2³·3)z−2²·3·2324·12)

Step 7

z²4−z6−(−2²·3·5z²+(2³·3)z−2²·3·2324·12)

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 2²·3.

z²4−z6−(2²·3(−2²·3·5z²2²·3+2³·3z2²·3−2²·3·232²·3)24·12)

Step 8

z²4−z6−(2²·3(−2²·3·5z²2²·3+2³·3z2²·3−2²·3·232²·3)24·12)

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:

2²·3.

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:

2²·3.

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:

2²·3.

z²4−z6−(2²·3(−1·1·5z²1·1+2³⁻²·1z1·1−1·1·231·1)24·12)

Step 9

z²4−z6−(2²·3(−1·1·5z²1·1+2³⁻²·1z1·1−1·1·231·1)24·12)

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.

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.

z²4−z6−(2²·3(−5z²1+2¹·z1−231)24·12)

Step 10

z²4−z6−(2²·3(−5z²1+2¹·z1−231)24·12)

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 5z².

This is a special case of exponentiation.

The following rule is applied:

A¹=A

In our example,

A 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 23.

z²4−z6−(2²·3(−(5z²)+2z1−23)24·12)

Step 11

z²4−z6−(2²·3(−(5z²)+2z1−23)24·12)

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:

5, z²

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 2z.

z²4−z6−(2²·3(−5z²+2z−23)24·12)

Step 12

z²4−z6−(2²·3(−5z²+2z−23)24·12)

We can reduce this fraction by dividing both numerator and denominator by a common numeric factors.

In our example,

both number 3 in numerator and number 24 in denominator are divisible by 3.

z²4−z6−(2²·1(−5z²+2z−23)8·12)

Step 13

z²4−z6−(2²·1(−5z²+2z−23)8·12)

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.

Let's multiply out the numbers after reduction.

In our example,

8·12 becomes 96

z²4−z6−(2²(−5z²+2z−23)96)

Step 14

z²4−z6−(2²(−5z²+2z−23)96)

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:

2²(−5z²+2z−23), 96

z²4−z6−2²(−5z²+2z−23)96

Step 15

z²4−z6−2²(−5z²+2z−23)96

We need to add fractions.

The following rule is applied:

AB+CD=LCDBA+LCDDCLCD

This example involves 3 terms.

The LCD is equal to:

4·6·96

(6·96)(z²)+(4·96)(−z)+(4·6)(−2²(−5z²+2z−23))4·6·96

Step 16

(6·96)(z²)+(4·96)(−z)+(4·6)(−2²(−5z²+2z−23))4·6·96

Numerical factors in this term have been multiplied.

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 factors in this term have been multiplied.

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.

576z²+384(−z)+24(−4(−5z²+2z−23))4·6·96

Step 17

576z²+384(−z)+24(−4(−5z²+2z−23))4·6·96

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.

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.

576z²−384z−24·4(−5z²+2z−23)4·6·96

Step 18

576z²−384z−24·4(−5z²+2z−23)4·6·96

Numerical factors in this term have been multiplied.

576z²−384z−96(−5z²+2z−23)4·6·96

Step 19

576z²−384z−96(−5z²+2z−23)4·6·96

We need to expand this term by multiplying 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 3 terms:

the first term is a product of 96 and −5z².

the second term is a product of 96 and 2z.

the third term is a product of 96 and −23.

576z²−384z−(−96·5z²+96·2z−96·23)4·6·96

Step 20

576z²−384z−(−96·5z²+96·2z−96·23)4·6·96

Numerical factors in this term have been multiplied.

576z²−384z−(−480z²+192z−2208)4·6·96

Step 21

576z²−384z−(−480z²+192z−2208)4·6·96

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 3 terms will change sign:

−480z², 192z, −2208

576z²−384z+480z²−192z+22084·6·96

Step 22

576z²−384z+480z²−192z+22084·6·96

We need to organize this expression into groups of like terms, so we can combine them easier.

There are 2 groups of like terms:

first group: 576z², 480z²

second group: −384z, −192z

576z²+480z²−384z−192z+22084·6·96

Step 23

576z²+480z²−384z−192z+22084·6·96

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 2 groups of like terms:

first group: 576z², 480z²

second group: −384z, −192z

1056z²−576z+22084·6·96

Step 24

1056z²−576z+22084·6·96

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.

Prime number 3 to the power of 1 equals 3.

Prime number 11 to the power of 1 equals 11.

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.

Prime number 3 to the power of 1 equals 3.

Prime number 23 to the power of 1 equals 23.

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 6 equals 64.

Prime number 3 to the power of 2 equals 9.

2⁵·3·11z²−(2⁶·3²)z+2⁵·3·234·6·96

Step 25

2⁵·3·11z²−(2⁶·3²)z+2⁵·3·234·6·96

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 2⁵·3.

2⁵·3(2⁵·3·11z²2⁵·3−2⁶·3²·z2⁵·3+2⁵·3·232⁵·3)4·6·96

Step 26

2⁵·3(2⁵·3·11z²2⁵·3−2⁶·3²·z2⁵·3+2⁵·3·232⁵·3)4·6·96

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:

2⁵·3.

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:

2⁵·3.

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:

2⁵·3.

2⁵·3(1·1·11z²1·1−2⁶⁻⁵·3²⁻¹·z1·1+1·1·231·1)4·6·96

Step 27

2⁵·3(1·1·11z²1·1−2⁶⁻⁵·3²⁻¹·z1·1+1·1·231·1)4·6·96

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.

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.

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.

2⁵·3(11z²1−2¹·3¹·z1+231)4·6·96

Step 28

2⁵·3(11z²1−2¹·3¹·z1+231)4·6·96

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 11z².

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 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 23.

2⁵·3(11z²−2·3z1+23)4·6·96

Step 29

2⁵·3(11z²−2·3z1+23)4·6·96

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 2·3z.

2⁵·3(11z²−(2·3z)+23)4·6·96

Step 30

2⁵·3(11z²−(2·3z)+23)4·6·96

Numerical factors in this term have been multiplied.

2⁵·3(11z²−(6z)+23)4·6·96

Step 31

2⁵·3(11z²−(6z)+23)4·6·96

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:

6, z

2⁵·3(11z²−6z+23)4·6·96

Step 32

2⁵·3(11z²−6z+23)4·6·96

We can reduce this fraction by dividing both numerator and denominator by a common numeric factors.

In our example,

both number 3 in numerator and number 6 in denominator are divisible by 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:

2⁵⁻¹·1(11z²−6z+23)4·1·96

Step 33

2⁵⁻¹·1(11z²−6z+23)4·1·96

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.

Let's multiply out the numbers after reduction.

In our example,

4·96 becomes 384

Note that since multiplying by 1 does not change the value of the product, factor 1 can be omitted.

2⁴(11z²−6z+23)384

Step 34

2⁴(11z²−6z+23)384

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.

16(11z²−6z+23)384

Step 35

16(11z²−6z+23)384

In order to reduce this fraction we first need to write numerical factors as product of primes.

In our example,

16 is rewritten as 2·2·2·2.

384 is rewritten as 2·2·2·2·2·2·2·3.

2·2·2·2(11z²−6z+23)2·2·2·2·2·2·2·3

Step 36

2·2·2·2(11z²−6z+23)2·2·2·2·2·2·2·3

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:

2·2·2·2.

1·1·1·1(11z²−6z+23)1·1·1·1·2·2·2·3

Step 37

1·1·1·1(11z²−6z+23)1·1·1·1·2·2·2·3

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 four times.

Let's multiply out the numbers after reduction.

In our example,

2·2·2·3 becomes 24

Note that since multiplying by 1 does not change the value of the product, factor 1 can be omitted.

11z²−6z+2324

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