Dividing Polynomials

According to the formula, the voltage of a circuit is found by dividing the power by the current. In this case, both the power and the current of the circuit are represented by polynomials. To find the voltage, these two polynomials have to be divided. Such division can be done following the steps of polynomial long division. First, write both polynomials in standard form, filling in any missing terms in the numerator with zeros. In $P(x),$ only the quadratic term is missing. Now, write the dividend $P(x)$ under the division symbol and the divisor $I(x)$ to the left. The first term of the quotient is found by dividing the leading term of $P(x)$ by the leading term of $I(x).$ Next, write the monomial above the horizontal bar. Multiply the divisor by $2x^2$ and write the resulting expression below the dividend. In this case, the multiplication gives $2x^4-8x^3+10x^2.$ Subtract the resulting polynomial from the dividend. Since the remainder has degree $3$ and the divisor has degree $2,$ the division is not done yet. Therefore, the process will be repeated until the remainder's degree is less than $2.$ The division is complete. The polynomial above the line is the quotient and the polynomial at the very bottom is the remainder. Finally, the voltage of the given circuit can be represented by the following expression.

Next, set up the coefficients of $P(x)$ and the synthetic division symbol. Here, $k=\text{-}2.$ Bring down the first coefficient, multiply it by $\text{-}2,$ and write the result in the second column. Add the numbers in the second column and write the result below the line. Then multiply the new number below the line by $\text{-}2$ and write the result in the third column. Next, add the numbers in the third column and write the result below the line. Multiply this number by $\text{-}2$ and write the result in the fourth column. It is almost done! Add the numbers in the fourth column and write the result below the line. Then, multiply this number by $\text{-}2$ and write the result in the fifth column. Finally, add the numbers in the last column and write the result below the line. The quotient of the division is formed by using the first four numbers below the line. The remainder is the very last number below the line. The quotient is one degree lower than the numerator. The procedure to perform the division is similar to the one applied in Part A. First, bring the number $3$ down. Next, multiply $4$ by $3$ and write the result in the second column and below $0.$ Next, find $0+12,$ which gives $12,$ and write the result in the same column but below the line. After that, multiply $4$ by $12$ and write the result in the third column and below $\text{-}4.$ As in the previous step, add the numbers in the third column. The sum of $\text{-}4$ and $48$ is $44,$ so write $44$ in the same column but below the line. Then multiply $\text{-}4$ by $44$ and write the resulting number in the fourth column and below $0.$ One more time, add the numbers in the fourth column and write the result in the same column but below the line. In this case, the sum is $176.$ Then, multiply $4$ by $176$ and write the result in the fifth column and below $\text{-}36.$ Finally, find the sum of the numbers in the last column and write the result below the line. The first four numbers below the line are the coefficients of the quotient and the fifth number is the remainder.

In the given case, the binomial is $x-3,$ which means that $k=3.$ Therefore, instead of performing the division, the remainder can be found simply by evaluating $P(x)$ at $x=3.$ Consequently, the remainder is $464$ — that is, $r=464.$ To verify that the result is correct, perform the polynomial long division. As shown, the remainder of the division is $464,$ as previously concluded. The remainder could also be found using synthetic division. Next, follow the steps of synthetic division. Bring down the number $2.$ Multiply $6$ by $2$ and write the result below $\text{-}13.$ Add the numbers in the second column and write the result below the line. The same steps are repeated over and over until the last column is reached. At the end of the process, it was concluded that $H(6)=0.$ Therefore, $6$ is a zero of $P(x).$ Consequently, by the Factor Theorem, $x-6$ is a factor of $H(x).$ In fact, $H(x)$ can be written as follows.