Evaluate a Polynomial Using Synthetic Substitution - Pre-Calculus

Recall that dividing a polynomial by does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example

where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that with a remainder of .

And how do we check to know if we have the right answer? We multiply and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3)

=

now we just add the remainder which is 1 to yield which matches the original dividend and is therefore our answer!

To determine if is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the being divided in. Combine the result of that with the next coefficient , which is . Then, multiply that by . Combine that result with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that does not go in evenly to this function and is not a root.

Synthetic division is a short cut for doing long division of polynomials and it can only be used when divifing by divisors of the form . The result or quoitient of such a division will either divide evenly or have a remainder. If there is no remainder, then the "" is said to be a factor of the polynomial. The polynomial must be in standard form (descending degree) and if a degree is skipped such as it must be accounted for by a "place holder".

         __ __ __

__ __        

where is the remainder.

While doing the long division we add vertically and we multiply diagonally by k. The empty lines represent places we put the sums and products. Notice that after the first term in the top row there is a 0; this is the place holder. This is because the degrees in the polynomial skipped. When the new coefficients have been found always rewrite starting with one order lower than the highest degree of the original polynomial.

Use synthetic division to verify each factor of the form . Lets start with .

Two goes into 6 three times resulting in:

From here we see will give you a remainder of zero and is therefore a factor of the polynomial .

Recall that dividing a polynomial by does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example

where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that with a remainder of .

And how do we check to know if we have the right answer? We multiply and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3)

=

now we just add the remainder which is 1 to yield which matches the original dividend and is therefore our answer!

To determine if is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the being divided in. Combine the result of that with the next coefficient , which is . Then, multiply that by . Combine that result with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that does not go in evenly to this function and is not a root.

Synthetic division is a short cut for doing long division of polynomials and it can only be used when divifing by divisors of the form . The result or quoitient of such a division will either divide evenly or have a remainder. If there is no remainder, then the "" is said to be a factor of the polynomial. The polynomial must be in standard form (descending degree) and if a degree is skipped such as it must be accounted for by a "place holder".

         __ __ __

__ __        

where is the remainder.

While doing the long division we add vertically and we multiply diagonally by k. The empty lines represent places we put the sums and products. Notice that after the first term in the top row there is a 0; this is the place holder. This is because the degrees in the polynomial skipped. When the new coefficients have been found always rewrite starting with one order lower than the highest degree of the original polynomial.

Use synthetic division to verify each factor of the form . Lets start with .

Two goes into 6 three times resulting in:

From here we see will give you a remainder of zero and is therefore a factor of the polynomial .

Recall that dividing a polynomial by does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example

where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that with a remainder of .

And how do we check to know if we have the right answer? We multiply and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3)

=

now we just add the remainder which is 1 to yield which matches the original dividend and is therefore our answer!

To determine if is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the being divided in. Combine the result of that with the next coefficient , which is . Then, multiply that by . Combine that result with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that does not go in evenly to this function and is not a root.

Synthetic division is a short cut for doing long division of polynomials and it can only be used when divifing by divisors of the form . The result or quoitient of such a division will either divide evenly or have a remainder. If there is no remainder, then the "" is said to be a factor of the polynomial. The polynomial must be in standard form (descending degree) and if a degree is skipped such as it must be accounted for by a "place holder".

         __ __ __

__ __        

where is the remainder.

While doing the long division we add vertically and we multiply diagonally by k. The empty lines represent places we put the sums and products. Notice that after the first term in the top row there is a 0; this is the place holder. This is because the degrees in the polynomial skipped. When the new coefficients have been found always rewrite starting with one order lower than the highest degree of the original polynomial.

Use synthetic division to verify each factor of the form . Lets start with .

Two goes into 6 three times resulting in:

From here we see will give you a remainder of zero and is therefore a factor of the polynomial .

Recall that dividing a polynomial by does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example

where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that with a remainder of .

And how do we check to know if we have the right answer? We multiply and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3)

=

now we just add the remainder which is 1 to yield which matches the original dividend and is therefore our answer!

To determine if is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the being divided in. Combine the result of that with the next coefficient , which is . Then, multiply that by . Combine that result with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that does not go in evenly to this function and is not a root.