Question

what is the highest degree for the expression (3x^2+4)(x^5-3)-12x^9-24x/3x+2

Answer 1

To find the highest degree of the expression \(\frac{(3x^2+4)(x^5-3)-12x^9-24x}{3x+2}\), we can break it down step by step.

1. **Calculate the degree of the numerator**:
- The first part of the numerator is \((3x^2 + 4)(x^5 - 3)\):
- The degree of \(3x^2 + 4\) is 2.
- The degree of \(x^5 - 3\) is 5.
- The degree of the product is \(2 + 5 = 7\).

Now we need to subtract \(12x^9 + 24x\) from this product:
- The degree of \(12x^9\) is 9.
- The degree of \(24x\) is 1.

Since \(12x^9\) has the highest degree (9), we need to consider this when subtracting:
- The leading term of \((3x^2 + 4)(x^5 - 3)\) is \(3x^2 \cdot x^5 = 3x^7\).
- We subtract \(12x^9\) from it, so the degree of the resulting polynomial is still determined by \(12x^9\) since \(9 > 7\).

Therefore, the overall degree of the numerator \((3x^2+4)(x^5-3) - 12x^9 - 24x\) is 9.

2. **Calculate the degree of the denominator**:
- The degree of the denominator \(3x + 2\) is 1.

3. **Calculate the degree of the entire expression**:
- The degree of the entire expression is given by the degree of the numerator minus the degree of the denominator:
\[
\text{Degree of the expression} = \text{Degree of numerator} - \text{Degree of denominator} = 9 - 1 = 8.
\]

Thus, the highest degree of the expression \(\frac{(3x^2+4)(x^5-3)-12x^9-24x}{3x+2}\) is \(8\).