Use long division to divide. Specify the quotient and the remainder.
(4x2 − 9x − 9) ÷ (4x + 3)

To solve \((4x^2 - 9x - 9) \div (4x + 3)\) using long division, let's proceed step by step.

### Step 1: Set up the division

We are dividing \(4x^2 - 9x - 9\) by \(4x + 3\).

\[
\frac{4x^2 - 9x - 9}{4x + 3}
\]

### Step 2: Divide the first term of the dividend by the first term of the divisor

Divide \(4x^2\) by \(4x\):
\[
\frac{4x^2}{4x} = x
\]
So, the first term in the quotient is \(x\).

### Step 3: Multiply and subtract

Now, multiply \(x\) by \(4x + 3\):
\[
x \cdot (4x + 3) = 4x^2 + 3x
\]

Subtract \((4x^2 + 3x)\) from \((4x^2 - 9x - 9)\):
\[
(4x^2 - 9x - 9) - (4x^2 + 3x) = -12x - 9
\]

### Step 4: Divide the new first term by the first term of the divisor

Divide \(-12x\) by \(4x\):
\[
\frac{-12x}{4x} = -3
\]
So, the next term in the quotient is \(-3\).

### Step 5: Multiply and subtract again

Now, multiply \(-3\) by \(4x + 3\):
\[
-3 \cdot (4x + 3) = -12x - 9
\]

Subtract \((-12x - 9)\) from \((-12x - 9)\):
\[
(-12x - 9) - (-12x - 9) = 0
\]

### Step 6: Conclusion

Since there is no remainder, the quotient is:
\[
x - 3
\]

Thus, the solution is:
\[
\boxed{(x - 3)} \quad \text{with a remainder of } 0.
\]

Would you like more details on any part? Here are some related questions:

1. How does synthetic division compare to long division in algebra?
2. How do you apply the remainder theorem in polynomial division?
3. Can you demonstrate how to divide polynomials with remainders?
4. What are some methods for factoring quadratic polynomials?
5. How do you handle dividing by more complex polynomials?

**Tip:** Always align terms based on the degree of the variable when performing polynomial division!

Use long division to divide. Specify the quotient and the remainder. (4x^2 − 9x − 9) ÷ (4x + 3)

Master the process of polynomial long division with this detailed step-by-step solution. Learn how to divide (4x^2 − 9x − 9) by (4x + 3), specify the quotient, and find the remainder. This guide is ideal for high school students in Grades 9-12.

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