In certain cases, you can factor the numerator to easily cancel out the divisor, making the division quick and straightforward.
Problem: Divide $6x^3 – x^2 – 10x – 3$ by $2x – 3$.
If we know the factors of the numerator:
Canceling the common term $(2x-3)$ gives the quotient:
This method shows the steps to factor the numerator by manipulating the terms to isolate the divisor $(2x – 3)$:
Now, substituting the factored form back into the division:
When a quick short cut isn’t available, Long Division is the reliable, systematic approach to finding the quotient and remainder.
Problem: Divide $10x^4 + 17x^3 – 62x^2 + 30x – 3$ by $2x^2 + 7x – 1$.
Here is the process, focusing on dividing the leading term of the remaining polynomial by the leading term of the divisor ($2x^2$) in each step:
| Step | Operation | Explanation |
| 1 | $\frac{10x^4}{2x^2} = 5x^2$ | First term of the quotient. Multiply the divisor by $5x^2$: $5x^2(2x^2 + 7x – 1) = 10x^4 + 35x^3 – 5x^2$. Subtract this from the dividend. |
| 2 | $\frac{-18x^3}{2x^2} = -9x$ | Second term of the quotient. Multiply the divisor by $-9x$: $-9x(2x^2 + 7x – 1) = -18x^3 – 63x^2 + 9x$. Subtract this from the remainder. |
| 3 | $\frac{6x^2}{2x^2} = 3$ | Third term of the quotient. Multiply the divisor by $3$: $3(2x^2 + 7x – 1) = 6x^2 + 21x – 3$. Subtract this from the remainder. |
The final subtraction results in 0, meaning the remainder is zero.
Final Result:
Since the remainder is zero, the dividend is exactly the product of the divisor and the quotient:
This method involves skillfully rewriting the terms of the numerator (dividend) to pull out the divisor $(2x^2 + 7x – 1)$ repeatedly.
Problem: Divide $10x^4 + 17x^3 – 62x^2 + 30x – 3$ by $2x^2 + 7x – 1$.
The goal is to factor the numerator:
Subtracting this from the original dividend leaves the remaining terms:
Note: The steps shown in the image are slightly different in arrangement but achieve the same result:
$$10x^4 + 17x^3 – 62x^2 + 30x – 3$$$$= 5x^2(2x^2 + 7x – 1) – 18x^3 – 57x^2 + 30x – 3$$(This is the remainder after the first step, grouped)
(This leaves a new, smaller remainder: $6x^2 + 21x – 3$)