Question

1. $(3a^4 + 4a^3 + 15a + 10) div (3a + 4)$

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(3a^{4}\) by the leading term of the divisor \(3a\). We get \(\frac{3a^{4}}{3a}=a^{3}\).
Multiply the divisor \((3a + 4)\) by \(a^{3}\): \(a^{3}(3a + 4)=3a^{4}+4a^{3}\).
Subtract this from the dividend: \((3a^{4}+4a^{3}+15a + 10)-(3a^{4}+4a^{3})=15a + 10\).

Step2: Now, divide the new leading term (but here we have a remainder situation? Wait, no, let's check again. Wait, the degree of the new dividend \(15a+10\) is less than the degree of the divisor \(3a + 4\), so we can express the division as:

\(\frac{3a^{4}+4a^{3}+15a + 10}{3a + 4}=\frac{3a^{4}+4a^{3}}{3a + 4}+\frac{15a + 10}{3a + 4}\)
We already found that \(\frac{3a^{4}+4a^{3}}{3a + 4}=a^{3}\), and for the second fraction, we can perform polynomial long - division or rewrite it. Let's do polynomial long - division for the second part. Divide \(15a\) by \(3a\) to get \(5\). Multiply \((3a + 4)\) by \(5\): \(5(3a + 4)=15a+20\). Subtract this from \(15a + 10\): \((15a + 10)-(15a + 20)=- 10\).
So, \(\frac{3a^{4}+4a^{3}+15a + 10}{3a + 4}=a^{3}+\frac{15a + 10}{3a + 4}=a^{3}+5-\frac{10}{3a + 4}\)

Wait, there is a mistake in the first step. Wait, the original dividend is \(3a^{4}+4a^{3}+15a + 10\), when we subtract \(3a^{4}+4a^{3}\) from it, we get \(15a + 10\). Now, let's do the division of \(15a+10\) by \(3a + 4\) correctly.
Let \(15a+10 = k(3a + 4)+r\), where \(k\) is the quotient and \(r\) is the remainder.
\(15a+10=3ka + 4k+r\)
Equating the coefficients of \(a\): \(15 = 3k\Rightarrow k = 5\)
Equating the constant terms: \(10=4k + r\). Substitute \(k = 5\), we get \(10=20 + r\Rightarrow r=- 10\)
So, \(\frac{15a + 10}{3a + 4}=5-\frac{10}{3a + 4}\)
And \(\frac{3a^{4}+4a^{3}}{3a + 4}=a^{3}\) (since \(3a^{4}+4a^{3}=a^{3}(3a + 4)\))

So, \(\frac{3a^{4}+4a^{3}+15a + 10}{3a + 4}=a^{3}+5-\frac{10}{3a + 4}\)

Answer:

\(a^{3}+5-\frac{10}{3a + 4}\)