Dividing Polynomials

According to the formula, we can find the current of a circuit by dividing the power by the voltage. Current = Power/Voltage In our case, both the power and the voltage of the circuit are represented by polynomials. P(x) &= 2x^4 - 20x^3 + 14x^2 - 72 V(x) &= 2x^2-4x+6 Let's divide these two polynomials to find the current of the circuit. We can do this by following the steps of polynomial long division. First, we write both polynomials in standard form, filling in any missing terms in the numerator with zeros. In P(x), only the linear term is missing. P(x) &= 2x^4 - 20x^3 + 14x^2 + 0x - 72 V(x) &= 2x^2-4x+6 Now, we write the dividend P(x) under the division symbol and the divisor V(x) to the left. 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 To find the first term of the quotient, we divide the leading term of P(x) by the leading term of V(x). 2x^4/2x^2 = x^2 Next, we write the resulting monomial above the horizontal bar. rr & x^2 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 After this, we multiply the divisor by x^2 and write the resulting expression below the dividend. In this case, the multiplication gives 2x^4-4x^3+6x^2. rr & x^2 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 & 2x^4 - 4x^3 + 6x^2 Let's subtract the resulting polynomial from the dividend. rr & x^2 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 & -( 2x^4-4x^3 + 6x^2) & -16x^3 + 8x^2 +0x-72 Since the result has degree 3 and the divisor has degree 2, the division is not done yet. Therefore, we will repeat the process until the result's degree is less than 2. rr & x^2 - 8x-12 2x^2-4x+6 & |r 2x^4 - 20x^3 + 14x^2 + 0x - 72 & -( 2x^4-4x^3 + 6x^2) & -16x^3 + 8x^2 +0x-72 & -( -16x^3 + 32x^2 - 48x) & -24x^2 +48x-72 & -( -24x^2-48x-72) & We completed the division. The polynomial above the line is the quotient and the polynomial at the very bottom is the remainder. Q(x) &= x^2 - 8x-12 R(x) &= 0 Since R(x)=0, the polynomial V(x) divides P(x) exactly. Therefore, we can write the following equation. 2x^4 - 20x^3 + 14x^2 - 72/2x^2-4x+6 = x^2 - 8x-12 From this, we conclude that the current of the given circuit is represented by the polynomial I(x)= x^2 - 8x-12.

Since the divisor is a binomial of the form x-k, we can perform the required division using synthetic division. The first thing we note is that P(x) has no quadratic term. Therefore, we fill in this term with a zero. 2x^4 - 4x^3 + 0x^2+ 12x - 24/x+3 Now, let's set up the coefficients of P(x) and the synthetic division symbol. Here, k= -3. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 Next, we bring down the first coefficient, multiply it by -3, and write the result inside the division symbol in the second column. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 &-6 & c 2 & The next step is to add the numbers in the second column and write the result below the line. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 & -6 & c 2 & -10 We repeat the steps through the last column. rl IR-0.15cm r -3 & |rr 2 & -4 & 0 & 12 & -24 & -6 & 30 & -90 & 234 & c 2 & -10 & 30 & -78 & 210 We are done! Remember, in synthetic division, the quotient's degree equals the degree of P(x) minus one. Therefore, the degree of Q(x) is 3. Additionally, the quotient is formed by using the first four numbers below the line. Q(x) = 2x^3 - 10x^2 + 30x - 78

As in Part A, we will perform the required division using synthetic division. The initial setup for the synthetic division is the same as in Part A except for the number to the left of the division symbol. Here, the divisor is x-2, therefore, k= 2. rl IR-0.15cm r 2 & |rr 2 & -4 & 0 & 12 & -24 Now, we bring down the first coefficient, multiply it by 2, and write the result in the second column above the horizontal line. After that, we add the numbers in the second column and write the result below the line. rl IR-0.15cm r 2 & |rr 2 & -4 & 0 & 12 & -24 & 4 & c 2 & 0 Let's repeat the previous steps until we reach the last column. rl IR-0.15cm r 2 & |rr 2 & -4 & 0 & 12 & -24 & 4 & 0 & 0 & 24 & c 2 & 0 & 0 & 12 & 0 We are done! Since the divisor has degree 1, the degree of the remainder is 0. Therefore, the remainder is a number. In fact, the remainder of the division is the rightmost number below the line. r = 0

We start by noticing that the divisors are linear binomials of the form x-k. Therefore, each division can be performed using synthetic division. However, since we are not asked for the quotients but only for the remainders, instead of applying the synthetic division three times, we can use the Remainder Theorem.

The Remainder Theorem |- If a polynomial P(x) is divided by a binomial of the form x-k, then the remainder of the division is equal to P(k).

This theorem tells us that we can find the remainder of each division simply by evaluating P(x) at the corresponding value. For instance, the remainder of P(x)÷ (x-1) is equal to P(1). Thus, let's find it.

Similarly, we can find the remainders of the other two divisions. The results are summarized in the table.

Now that we know the three remainders, let's add them up. Sum &= 8 + 14 + 653 &⇓ Sum &= 675 ✓

Let's begin by verifying that Heichi's note is incorrect. We can do it by using integer numbers — after all, integers are 0-degree polynomials. For example, let's consider the division 11 ÷ 4. & 2 4) & l 11 & -8 & 3 The quotient of the division is 2 and the remainder is 3. Now, let's substitute these values into the equation that Heichi wrote.

As seen, the left-hand side is not equal to the right-hand side. The same will happen for any polynomial division. To write the correct statement, notice that in the previous computations, the right-hand side resulted to be equal to the numerator. That is, the right-hand side in Heichi's note is equal to P(x). P(x) = Q(x)D(x) + R(x) The previous equation is true for integers and also for polynomials. Now, let's divide both sides of this equation by D(x) and simplify.

From the above, the correct right-hand side is equal to the quotient of the division plus the division of the remainder and the divisor. P(x)/D(x) = Q(x) + R(x)/D(x) We can test this equation with the numbers we used before.

As seen, we got a true statement and this is true for any polynomial division.

When we divide a polynomial P(x) by another polynomial D(x), we get a quotient Q(x) and a remainder R(x). P(x)/D(x) → & Q(x) D(x) & | l P(x) & ... & R(x) Moreover, once we know the quotient and remainder, we can write the original division as the quotient plus the division of the remainder and the divisor. P(x)/D(x) = Q(x) + R(x)/D(x) In the given situation, the divisor, quotient, and remainder are known. Thus, to find the dividend, we can solve the previous equation for P(x) and then substitute the corresponding polynomials.

We are told that the divisor is x-2, therefore D(x)=x-2. Similarly, Q(x)=3x^3-2x+7 and R(x)=-3. Let's substitute these expressions into the last equation to finally get P(x).