Problem 14 - Precalculus Mathematics for Calculus

Question

Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express the quotient \(P(x) / D(x)\) in the form $$\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$$ $$P(x)=x^{5}+x^{4}-2 x^{3}+x+1, \quad D(x)=x^{2}+x-1$$

Step-by-Step Solution

Answer

\(Q(x) = x^3 - x + 1\), remainder \(-x + 2\).

Step 1: Set Up the Long Division

Write the dividend \(P(x) = x^5 + x^4 - 2x^3 + x + 1\) under the long division symbol and the divisor \(D(x) = x^2 + x - 1\) outside. Align the terms according to their degrees, filling any missing terms with zero coefficients as placeholders.

Step 2: Divide Leading Terms

Divide the leading term of the dividend, \(x^5\), by the leading term of the divisor, \(x^2\). This gives \(x^3\). Write \(x^3\) above the division line as the first term of the quotient \(Q(x)\).

Step 3: Multiply and Subtract

Multiply the entire divisor \(D(x) = x^2 + x - 1\) by \(x^3\), giving \(x^5 + x^4 - x^3\). Subtract this from the current dividend \(x^5 + x^4 - 2x^3 + x + 1\) to get a new dividend \(0x^5 + 0x^4 - x^3 + x + 1\).

Step 4: Repeat Division Process

With the new dividend \(-x^3 + x + 1\), divide \(-x^3\) by \(x^2\), which equals \(-x\). Write \(-x\) in the quotient. Multiply \(-x\) by \(x^2 + x - 1\) to get \(-x^3 - x^2 + x\) and subtract this from \(-x^3 + x + 1\) to get \(0x^3 + x^2 + 1\).

Step 5: Final Division

Now divide \(x^2\) by \(x^2\), which results in \(1\). Write \(1\) as the last term of the quotient. Multiply \(1\) by \(x^2 + x - 1\), resulting in \(x^2 + x - 1\). Subtract from \(x^2 + 1\) to get a remainder of \(-x + 2\).

Step 6: Formulate Final Answer

The quotient \(Q(x) = x^3 - x + 1\) and the remainder \(R(x) = -x + 2\). Thus, \(\frac{P(x)}{D(x)} = x^3 - x + 1 + \frac{-x + 2}{x^2 + x - 1}\).

Key Concepts

Synthetic Division

Synthetic Division is a simplified method for dividing polynomials, particularly useful when the divisor is of the form \(x - a\). Unlike traditional long division, synthetic division requires fewer steps and computations, streamlining the process of finding the quotient and remainder.
To perform synthetic division, follow these basic steps:

• Identify \(a\) from your divisor \(x - a\).
• Write down the coefficients of the dividend polynomial.
• Bring down the leading coefficient to the bottom row.
• Multiply \(a\) by the value just written at the bottom, and write the result under the next coefficient.
• Add the result to the next coefficient, writing the sum on the bottom row. Repeat these steps for each coefficient.

The final row gives the quotient coefficients, and the last number is the remainder. Simple, right? This method cleverly captures the essence of polynomial division in an efficient way.
However, remember that synthetic division only works directly with linear divisors like \(x - a\). For more complex divisors, other methods such as long division or factoring may be required.

Long Division

Long Division is the go-to method when dealing with polynomial division, regardless of the complexity of the divisor. It allows you to divide any polynomial by another, yielding both a quotient and remainder. While it may seem complex initially, each step follows a logical sequence.
Here is how you perform long division on polynomials:

• Divide the leading term of the dividend by the leading term of the divisor to get the first term of the quotient.
• Multiply the entire divisor by this quotient, and write the result below the dividend.
• Subtract this result from the dividend to get a new polynomial expression.
• Repeat the process with this new expression until the degree of the remainder is less than the degree of the divisor.

Let's take a closer examination of our problem. When dividing \(P(x) = x^5 + x^4 - 2x^3 + x + 1\) by \(D(x) = x^2 + x - 1\), we take it term by term, carefully applying the multiplication and subtraction until our remainder is smaller in degree than the divisor. This method ensures comprehensive division across all polynomial terms, providing clarity and precision in reaching the final quotient and remainder.

Remainder Theorem

The Remainder Theorem offers a quick way to find the remainder of a polynomial division when dividing by a linear divisor \(x - a\). This theorem beautifully links the remainder to a very simple calculation, sparing us from dividing large polynomials when not needed!
The Remainder Theorem states that for a polynomial \(P(x)\), when divided by \(x - a\), the remainder is \(P(a)\). This means:

• If you need just the remainder when dividing by \(x - a\), simply evaluate the polynomial at \(a\).
• There is no need to compute the entire division if the quotient is not required.

For example, if we had our polynomial \(P(x) = x^5 + x^4 - 2x^3 + x + 1\) and divisor \(x - 1\), then the remainder is \(P(1)\). Just plug \(1\) into the polynomial to find it. This method is not only time-efficient but also invaluable when time constraints or simplicity is sought, showcasing the genius of polynomial mathematics.