Step-by-step polynomial long division. Check with remainder theorem.

Divide 2x³ + x² - 5x + 2 by 2x - 1 — Polynomial Long Division

Question

Divide $2x^3 + x^2 - 5x + 2$ by $2x - 1$ and find the quotient and remainder.

Solution — Step by Step

Write the dividend $2x^3 + x^2 - 5x + 2$ and divisor $2x - 1$ in standard long division form. Both are already in descending order of degree — good, we’re ready.

Ask: what do we multiply $2x$ by to get $2x^3$ ? The answer is $x^2$ .

Multiply the full divisor: $x^2 \cdot (2x - 1) = 2x^3 - x^2$ .

Subtract from the dividend:

(2x^3 + x^2) - (2x^3 - x^2) = 2x^2

Bring down $-5x$ to get $2x^2 - 5x$ .

What do we multiply $2x$ by to get $2x^2$ ? That’s $x$ .

Multiply: $x \cdot (2x - 1) = 2x^2 - x$ .

Subtract:

(2x^2 - 5x) - (2x^2 - x) = -4x

Bring down $+2$ to get $-4x + 2$ .

What do we multiply $2x$ by to get $-4x$ ? That’s $-2$ .

Multiply: $-2 \cdot (2x - 1) = -4x + 2$ .

Subtract:

(-4x + 2) - (-4x + 2) = 0

Remainder is 0.

2x^3 + x^2 - 5x + 2 = (2x - 1)(x^2 + x - 2) + 0

Quotient: $x^2 + x - 2$ , Remainder: $0$

Since the remainder is 0, $(2x - 1)$ is a factor of the polynomial.

Why This Works

Polynomial long division mirrors integer long division exactly. We always divide the leading term of the current remainder by the leading term of the divisor — this gives us the next term of the quotient. The degree drops by 1 each round, which is why we stop when the remainder’s degree is less than the divisor’s degree.

Here, the divisor $2x - 1$ has degree 1, so we stop when our remainder is a constant (degree 0) or zero. Three rounds for a cubic dividend divided by a linear divisor — that’s always how it works: degree of quotient = degree of dividend − degree of divisor = $3 - 1 = 2$ .

The zero remainder tells us something extra: $x = \frac{1}{2}$ is a root of $2x^3 + x^2 - 5x + 2$ . Useful if this question had asked us to factorise.

Alternative Method — Verify with Remainder Theorem

The Remainder Theorem says: when $p(x)$ is divided by $(2x - 1)$ , the remainder equals $p\!\left(\frac{1}{2}\right)$ .

p\!\left(\tfrac{1}{2}\right) = 2\!\left(\tfrac{1}{2}\right)^3 + \left(\tfrac{1}{2}\right)^2 - 5\!\left(\tfrac{1}{2}\right) + 2

= 2 \cdot \tfrac{1}{8} + \tfrac{1}{4} - \tfrac{5}{2} + 2 = \tfrac{1}{4} + \tfrac{1}{4} - \tfrac{5}{2} + 2 = \tfrac{1}{2} - \tfrac{5}{2} + 2 = -2 + 2 = 0

Remainder $= 0$ ✓ — matches what long division gave us.

Always use the Remainder Theorem as a 30-second check after long division. Substitute $x = \frac{c}{d}$ (where the divisor is $dx - c$ ) into $p(x)$ . If it matches your remainder, the division is correct. This saves you from carrying an arithmetic error all the way to the end in board exams.

We can also factor the quotient further: $x^2 + x - 2 = (x + 2)(x - 1)$ .

So the full factorisation is:

2x^3 + x^2 - 5x + 2 = (2x - 1)(x + 2)(x - 1)

Common Mistake

The most common error is a sign mistake during subtraction. When we subtract $2x^3 - x^2$ from $2x^3 + x^2$ , many students write $+x^2 - x^2 = 0$ instead of $+x^2 - (-x^2) = 2x^2$ . The negative sign distributes to both terms of the product you’re subtracting. A clean way to avoid this: write the subtraction explicitly as adding the negative — change the sign of every term in the row you’re subtracting, then add.

Question

Solution — Step by Step

Why This Works

Alternative Method — Verify with Remainder Theorem

Common Mistake

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