Problem 87 - Algebra 2

Question

Divide. Tell whether each divisor is a factor of the dividend. $$ \left(6 a^{3}+a^{2}-a+4\right) \div(2 a+1) $$

Step-by-Step Solution

Answer

The quotient after dividing \(6a^3+a^2-a+4\) by \(2a+1\) is \(3a^2-a\) and the remainder is 4. Since there is a non-zero remainder, \(2a+1\) is not a factor of \(6a^3+a^2-a+4\).

Step 1: Set up the Division

We set up the expression in the form of long division by placing the dividend (the polynomial we are dividing) under the division bar, and the divisor outside to its left. So, our expression will be set up like: \[ (2a+1) \div (6 a^{3}+a^{2}-a+4) \]

Step 2: Divide the First Terms

Start by dividing the first term of the dividend (\(6a^3\)) by the first term of the divisor (\(2a\)). The result is \(3a^2\), which you write above the division bar.

Step 3: Multiply and Subtract

Now, you multiply the divisor (\(2a+1\)) by the term you just wrote above the division bar (\(3a^2\)). You get \(6a^3+3a^2\). Write this underneath the first two terms of the dividend and subtract it from those terms. After subtraction, it simplifies to \( -2a^2-a+4 \). Write it down.

Step 4: Repeat Division Process

Continue the process started in the second step with the result from step 3 (\( -2a^2-a+4 \)) as your new dividend and repeat the process. The first term (\( -2a^2\)) divided by the first term of the divisor (\(2a\)) will give \(-a\). Multiply the divisor by \(-a\), giving \(-2a^2 - a\). This is subtracted from \(-2a^2 - a + 4\), leaving you with 4.

Step 5: Final Division and Conclude

Finally, the process is repeated with 4 as the new dividend. But as 4 cannot be divided by \(2a\), it remains as it is. This is the remainder and since it is non-zero, we can infer that \(2a+1\) is not a factor of \(6a^3+a^2-a+4\). Hence, the quotient of the division is \(3a^2-a\) and the remainder is 4.

Key Concepts

Divisibility

In mathematics, divisibility is a concept that determines whether one number is a factor of another. When dealing with polynomials, divisibility helps us understand if a given polynomial can be perfectly divided by another without leaving a remainder. If a polynomial is divisible by another, it means that the division yields a quotient with a remainder of zero.

• If the remainder is zero, the divisor is considered a factor of the dividend.
• If the remainder is not zero, the divisor is not a factor.

In the context of our polynomial division problem, we checked if \(2a + 1\) is a factor of \(6a^3 + a^2 - a + 4\). After completing the division process and discovering a non-zero remainder (which is 4), we concluded that \(2a + 1\) is not a factor.

Remainder Theorem

The Remainder Theorem is a valuable tool in polynomial division. It provides a shortcut to determine the remainder of a polynomial divided by a binomial of the form \(x - c\). According to the theorem, if you divide a polynomial \(f(x)\) by \(x - c\), the remainder of that division is simply \(f(c)\).

• This theorem can save you from performing lengthy polynomial division steps.
• It is especially useful when checking if a polynomial is divisible by a particular divisor.

Although the exercise we worked on involves a division by a binomial not in the form \(x - c\), the Remainder Theorem still underscores the importance of remainders in verifying divisibility.

Synthetic Division

Synthetic division is a simpler method for dividing a polynomial by a binomial of the form \(x - c\). It is an efficient and quick alternative to long division for such problems. It uses less writing and fewer calculations than traditional polynomial long division.

• Synthetic division typically involves just numbers and not variables.
• It is most effective when the divisor is a straightforward binomial.

In the division problem we solved, we followed the long division method. However, if our divisor could be structured in a suitable way for synthetic division, we would likely have employed this method for faster computation.

Factor Theorem

The Factor Theorem is closely linked to both the Remainder Theorem and divisibility. It states that a polynomial has a factor \(x - c\) if and only if \(f(c) = 0\) where \(f(x)\) is the polynomial. This simplifies finding factors since:

• If \(f(c) = 0\), then \(x - c\) is a factor of \(f(x)\).
• If \(f(c) eq 0\), then \(x - c\) is not a factor.

While solving the exercise, we checked if \(2a + 1\) was a factor of \(6a^3 + a^2 - a + 4\). The non-zero remainder of 4 confirmed that \(2a + 1\) was not a factor, which aligns with the principles of the Factor Theorem. This method is beneficial for quickly determining if any given binomial is a factor of a polynomial.