Problem 69 - Elementary and Intermediate Algebra

Question

Perform each division. $$ \frac{3 b^{2}+11 b+6}{3 b+2} $$

Step-by-Step Solution

Answer

The quotient is \(b + 3\).

Step 1: Setup the Division

We start by writing the given division as a long division problem: \[\frac{3b^2 + 11b + 6}{3b + 2}\] We are dividing the polynomial \(3b^2 + 11b + 6\) by \(3b + 2\). This means 3b+2 is our divisor, and 3b^2 + 11b + 6 is our dividend.

Step 2: Divide the Leading Terms

Look at the leading terms of the divisor (3b) and the dividend (3b^2). Divide these to find the first term of the quotient:\[\frac{3b^2}{3b} = b\]This means that \(b\) is the first term of our quotient.

Step 3: Multiply and Subtract

Multiply \(b\) by the entire divisor (3b + 2):\[b imes (3b + 2) = 3b^2 + 2b\]Subtract this product from the dividend:\[(3b^2 + 11b + 6) - (3b^2 + 2b) = 9b + 6\]

Step 4: Repeat the Process

Now, treat \(9b + 6\) as the new dividend. Divide the leading term 9b by the leading term of the divisor 3b:\[\frac{9b}{3b} = 3\]So, 3 is the second term of the quotient.

Step 5: Final Multiply and Subtract

Multiply 3 by the divisor (3b + 2):\[3 imes (3b + 2) = 9b + 6\]Subtract this from the current dividend:\[(9b + 6) - (9b + 6) = 0\]Since the remainder is 0, the division is exact.

Step 6: Write the Final Answer

Since we have no remainder, the complete division gives us the quotient:\[b + 3\]

Key Concepts

Long Division of Polynomials

When you first encounter polynomial division, it can seem a bit overwhelming, but it's quite similar to the long division of numbers you might have learned in elementary school. In our problem, we're dividing a polynomial, \(3b^2 + 11b + 6\), by another polynomial, \(3b + 2\). Here’s how you perform long division with polynomials:

• Set Up the Division: Write the division in long division format, where the divisor \(3b + 2\) is placed outside the division symbol and the dividend \(3b^2 + 11b + 6\) is placed inside.
• Identify Leading Terms: Focus on the leading term in both the dividend and the divisor, which are \(3b^2\) and \(3b\) respectively in this case.
• Divide and Multiply: Divide the leading term of the dividend \(3b^2\) by the leading term of the divisor \(3b\), giving you \(b\). Then multiply the entire divisor by \(b\) and subtract from the dividend.
• Repeat the Process: Use the result of your subtraction as the new dividend and repeat the steps, identifying leading terms in the new dividend.

After repeating the division process, you eventually find yourself with a remainder, or in this particular case, zero. This iterative process is the core of long division of polynomials.

Division Algorithm

The division algorithm is fundamental to understanding polynomial division. The technique is similar to dividing numbers but involves a more abstract set of principles. It states that any polynomial dividend \(f(x)\) can be expressed as:\[ f(x) = d(x)q(x) + r(x) \]where:

• \(d(x)\) is the divisor, in our example \(3b + 2\).
• \(q(x)\) is the quotient you arrive at through the division. For the given exercise, it’s \(b + 3\).
• \(r(x)\) is the remainder, which is zero if the division is exact.

In our exercise, once the calculation is completed, you end up with a remainder of zero, meaning the polynomial \(3b^2 + 11b + 6\) can be exactly expressed by multiplying \(3b + 2\) by \(b + 3\), confirming the division algorithm. Understanding this concept deeply aids in simplifying polynomial expressions and finding roots.

Remainder Theorem

The remainder theorem offers a quick way to resolve whether your division process is complete and exact. According to this theorem, when dividing a polynomial \(f(x)\) by a linear divisor \(x-a\), the remainder of the division is \(f(a)\). If the remainder is zero, it indicates that \(x-a\) is a factor of \(f(x)\).For the division example \(\frac{3b^2 + 11b + 6}{3b + 2}\), we found that the remainder is zero. This outcome indicates not just a clean division but also tells us that \(3b + 2\) is a factor of \(3b^2 + 11b + 6\). Therefore, when you solve polynomial equations or factor them, checking the remainder quickly informs you about factorization feasibility without fully completing long division. The theorem thus streamlines polynomial factorization and simplifies solving polynomial equations.