What is the first step of the division problem (8x^3 - x^2 + 6x + 7) divided by (2x - 1)? Choices: divide 8x^3 by 2x; divide 2x by 8x^3; divide 6x by 2x; divide 2x by 6x.

Step-by-step solution

1. 1

   Recall the rule that drives long division. At every stage you divide the leading term of what is left of the dividend by the leading term of the divisor. This is what guarantees the degree drops on each pass, so the process terminates.

2. 2

   Identify the two leading terms. In $8x^3-x^2+6x+7$ the leading term is $8x^3$ (highest power). In $2x-1$ the leading term is $2x$. So the first step divides $8x^3$ by $2x$.

3. 3

   Carry out that division. $\frac{8x^3}{2x}=4x^2$, which becomes the first term of the quotient. Multiplying back gives $4x^2(2x-1)=8x^3-4x^2$, and subtracting leaves $3x^2+6x+7$.

4. 4

   See why the other options fail. Dividing $2x$ by $8x^3$ reverses dividend and divisor and would give a negative power. Using $6x$ instead of $8x^3$ picks a non-leading term, which does not reduce the degree of the remainder.

5. 5

   Finish the division to confirm the setup works. Continuing gives quotient $4x^2+\frac32 x+\frac{15}{4}$ with remainder $\frac{43}{4}$; the fractional coefficients are fine, but the process only closes because each step used the leading terms.

6. 6

   Numerical check at x = 1. Dividend $=8-1+6+7=20$ and divisor $=1$, so the quotient value must be $20$; and $4+\frac32+\frac{15}{4}+\frac{43}{4}=20$, matching exactly.