Find the quotient (8x^3 - 6x^2 - 5x + 3) divided by (2x - 5).

Step-by-step solution

1. 1

   Set up with a non-monic divisor. Since $2x-5$ leads with $2x$, each quotient term is obtained by dividing the current leading term by $2x$. The dividend has all four powers present, so nothing needs a placeholder.

2. 2

   First step. $\dfrac{8x^3}{2x}=4x^2$. Multiply back: $4x^2(2x-5)=8x^3-20x^2$. Subtract: $-6x^2-(-20x^2)=14x^2$, bring down $-5x$ to get $14x^2-5x$.

3. 3

   Second step. $\dfrac{14x^2}{2x}=7x$. Multiply back: $7x(2x-5)=14x^2-35x$. Subtract: $-5x-(-35x)=30x$, bring down $+3$ to get $30x+3$.

4. 4

   Third step. $\dfrac{30x}{2x}=15$. Multiply back: $15(2x-5)=30x-75$. Subtract: $3-(-75)=78$, a constant, so the division stops here.

5. 5

   State the result. $\dfrac{8x^3-6x^2-5x+3}{2x-5}=4x^2+7x+15+\dfrac{78}{2x-5}$; the remainder $78\neq 0$ shows $2x-5$ is not a factor.

6. 6

   Check both ways. Multiplying back: $(2x-5)(4x^2+7x+15)+78=8x^3+14x^2+30x-20x^2-35x-75+78=8x^3-6x^2-5x+3$. Evaluating the dividend at the divisor's root $x=\tfrac52$ gives $8\cdot\tfrac{125}{8}-6\cdot\tfrac{25}{4}-\tfrac{25}{2}+3=125-37.5-12.5+3=78$, matching the remainder.