Problem 26 - Precalculus: Functions and Graphs

Question

Use synthetic division to find the quotient and remainder If the first polynomial is divided by the second. $$-2 x^{4}+10 x-3 ; \quad x-3$$

Step-by-Step Solution

Answer

Quotient: \(-2x^3 - 6x^2 - 18x - 44\), Remainder: \(-135\).

Step 1: Set up the synthetic division

Write down the coefficients of the polynomial \(-2x^4 + 0x^3 + 0x^2 + 10x - 3\). The coefficients are \(-2, 0, 0, 10, -3\). Place 3, the zero of \(x - 3\), to the left.

Step 2: Bring down the leading coefficient

Bring down the first coefficient \(-2\) to start the synthetic division process. This coefficient will remain unchanged.

Step 3: Multiply and add

Multiply the number brought down by 3 (the zero of \(x-3\)), which gives \(-2 \times 3 = -6\). Write \(-6\) under the next coefficient, 0, and add, resulting in \(0 + (-6) = -6\).

Step 4: Repeat the multiply and add

Continue the process: multiply \(-6\) by 3 to get \(-18\), then add to the next coefficient, 0, resulting in \(0 + (-18) = -18\).

Step 5: Continue with synthetic division

Multiply \(-18\) by 3 to get \(-54\), and add to the next coefficient, 10, resulting in \(10 + (-54) = -44\).

Step 6: Finalize the synthetic division

Multiply \(-44\) by 3 to get \(-132\) and add to the final coefficient, \(-3\), which gives \(-3 + (-132) = -135\).

Step 7: Interpret the result

The synthetic division results in the coefficients \(-2, -6, -18, -44\) with a remainder of \(-135\). The quotient is \(-2x^3 - 6x^2 - 18x - 44\) and the remainder is \(-135\).

Key Concepts

Quotient and Remainder

When performing polynomial division, similar to regular division, you end up with a quotient and a remainder. These are the key results you're looking for.

The **quotient** is what you get when you divide the polynomial by another polynomial until it can't be divided further. In our exercise, when dividing \(-2x^4 + 10x - 3\) by \(x-3\), the quotient was determined to be \(-2x^3 - 6x^2 - 18x - 44\). This essentially means that if you multiply the quotient by the divisor \((x-3)\) and then add the remainder, you'll return to the original polynomial.

The **remainder** is what's left over after you've completed the division. Sometimes, this might be zero, especially if the dividend is exactly divisible by the divisor. In our example, the remainder was \(-135\). Even though it's considered a small value, it plays an important role in giving us a complete answer for our division problem.

Polynomial Division

Polynomial division might sound complex, but it's really just a structured way of simplifying bigger equations into more manageable pieces. Synthetic division is a method used to divide a polynomial by a linear factor like \((x - c)\).

Here are some features of synthetic division:

• **Simplification**: By using synthetic division, we avoid long polynomial division which can be cumbersome and time-consuming.
• **Efficiency**: Because it only involves addition, subtraction, and multiplication, it's a quicker method than traditional polynomial division.

During this process, placeholders are used for any missing terms. In our equation, terms like \(0x^3\) and \(0x^2\) were used to ensure all power levels of \(x\) are considered.

Synthetic division is particularly useful when the divisor is of the form \(x - c\). By knowing this, it should encourage you to look for opportunities to apply it where possible.

Precalculus Mathematics

Precalculus provides foundational knowledge needed to explore advanced mathematical concepts. In exercises like these, understanding polynomial division is vital for solving higher-level mathematics problems.

Precalculus is crucial because:

• **Foundation**: It lays the groundwork for calculus, including operations like division, manipulating expressions, and solving equations.
• **Application**: It allows for the analysis and interpretation of more complex polynomial relationships, which are important in calculus.

By mastering concepts such as synthetic division in precalculus, students build their analytical skills. These exercises also familiarize students with problem-solving tactics that they will use in calculus and beyond.

This foundation is not just about computation—it's about developing a deeper understanding of mathematical relationships and structures.