#### Answer

$b^2-3b-1+\frac{3}{3b-1}$

#### Work Step by Step

In order to divide a polynomial by a binomial, there must be a term for every power between the highest power and zero. To do this, add a placeholder of $0b$
$$(3b^3-10b^2+0b+4)\div(3b-1)$$
In order to match the first term of the dividend, multiply the divisor by $b^2$. The $b^2$ will go on top of your division sign.
$$(b^2)(3b-1)=3b^3-b^2$$
Subtract this from the first two terms of the dividend
$$(3b^3-10b^2+0b+4)-(3b^3-b^2)=-9b^2$$
Bring down the next term of the dividend
$$-9b^2+0b$$
Multiply the divisor by $-3b$ to match this. The $-3b$ will follow the $b^2$ on top of your division sign
$$(-3b)(3b-1)=-9b^2+3b$$
Subtract this from the dividend
$$(-9b^2+0t)-(-9b^2+3b)=-3b$$
Bring down the next term of the dividend
$$-3b+4$$
Multiply the divisor by $-1$ to match this. The $-1$ will follow the $-3b$ on top of your division sign
$$(-1)(3b-1)=-3b+1$$
Subtract this from the dividend
$$(-3b+4)-(-3b+1)=3$$
Since your remainder is 3, you can add $\frac{3}{3b-1}$ to your answer
$$b^2-3b-1+\frac{3}{3b-1}$$