Problem 4 - Algebra and Trigonometry

Question

Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express \(P\) in the form \(P(x)=D(x) \cdot Q(x)+R(x)\). \(P(x)=4 x^{3}+7 x+9, \quad D(x)=2 x+1\)

Step-by-Step Solution

Answer

The polynomial division yields: \(P(x) = (2x + 1)(2x^2 - x + 4) + 5\).

Step 1: Set Up the Long Division

First, set up the long division by writing down the dividend, which is the polynomial \(P(x) = 4x^3 + 0x^2 + 7x + 9\), and the divisor \(D(x) = 2x + 1\). Add a placeholder for the missing \(x^2\) term.

Step 2: Divide the Leading Terms

Divide the leading term of the dividend \(4x^3\) by the leading term of the divisor \(2x\), which gives \(2x^2\). This will be the first term of the quotient \(Q(x)\).

Step 3: Multiply and Subtract

Multiply \(2x^2\) by the entire divisor \(2x + 1\) to get \(4x^3 + 2x^2\). Subtract \(4x^3 + 2x^2\) from \(4x^3 + 0x^2 + 7x + 9\) to get \(-2x^2 + 7x + 9\).

Step 4: Repeat the Division Process

Divide the new leading term \(-2x^2\) by \(2x\), resulting in \(-x\). Multiply \(-x\) by \(2x + 1\) to get \(-2x^2 - x\). Subtract this from \(-2x^2 + 7x + 9\) to get \(8x + 9\).

Step 5: Final Division Step

Divide \(8x\) by \(2x\) to get \(4\). Multiply \(4\) by \(2x + 1\) to get \(8x + 4\). Subtract this from \(8x + 9\) to obtain the remainder \(5\).

Step 6: Express the Result

The quotient \(Q(x)\) is \(2x^2 - x + 4\) and the remainder \(R(x)\) is \(5\). Therefore, express \(P(x)\) as \(P(x) = (2x + 1)(2x^2 - x + 4) + 5\).

Key Concepts

Synthetic Division

Synthetic Division is a simplified method for dividing a polynomial by a binomial of the form \(x - c\). Unlike long division, it uses only the coefficients from the polynomials, making it quicker and easier for some calculations.
To start with synthetic division, list the coefficients of the polynomial \(P(x)\). In our example, \(P(x) = 4x^3 + 7x + 9\) has the coefficients \([4, 0, 7, 9]\). The divisor \(D(x) = 2x + 1\) can be rewritten as \(x - (-\frac{1}{2})\).
Once the setup is complete, follow these straightforward steps:

• Bring down the first coefficient.
• Multiply the divisor's value \((-\frac{1}{2})\) by the coefficient, then add to the next coefficient.
• Continue this pattern across all coefficients.

By the end, you will achieve the quotient and the remainder. Synthetic division is particularly valuable for polynomials where the divisor is simple, saving steps and reducing complexity.

Long Division Technique

The Long Division Technique is a traditional method for dividing polynomials, particularly useful for complex divisors or when the divisor does not fit the simple form required for synthetic division.
To perform long division, arrange the dividend \(P(x) = 4x^3 + 7x + 9\) in descending powers of \(x\), including placeholders for any missing terms, such as the \(0x^2\) included here. The divisor \(D(x) = 2x + 1\) goes on the left.
Here’s how to proceed:

• Divide the leading term of the dividend by the leading term of the divisor to get the first part of the quotient.
• Multiply this part of the quotient by the full divisor and subtract the result from the dividend.
• Bring down the next term of the dividend and repeat the process.

Keep repeating the division until reaching a term smaller than the divisor. This term becomes the remainder when no further division can be done. Long division gives a clear view of how the divisor fits into the dividend.

Quotient and Remainder

When you divide polynomials, you'll often express the result as a combination of the Quotient and Remainder.
The quotient \(Q(x)\) is what you get from dividing the polynomials step by step; in our example, it is \(2x^2 - x + 4\).
The remainder \(R(x)\) is what’s left over after the division is complete. In this case, the remainder is \(5\).
Thus, the division process can be represented by the equation:

• \(P(x) = D(x) \cdot Q(x) + R(x)\)

This expression allows you to see how \(P(x)\) is composed of multiples of \(D(x)\) plus a remainder. Understanding this is crucial as it helps in many areas of algebra and can simplify more advanced operations, such as solving polynomial equations and factoring.