## Algebra 1

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