Divide p(x) = x⁴-3x²+4x+5 by g(x) = x²+1-x

Question

Divide $p(x) = x^4 - 3x^2 + 4x + 5$ by $g(x) = x^2 + 1 - x$ .

Find the quotient $q(x)$ and remainder $r(x)$ , and verify using the division algorithm.

Solution — Step by Step

Before dividing, we write both polynomials with terms in descending order of degree:

Dividend: $p(x) = x^4 + 0 \cdot x^3 - 3x^2 + 4x + 5$
Divisor: $g(x) = x^2 - x + 1$

Note that we re-ordered $g(x) = x^2 + 1 - x$ as $x^2 - x + 1$ . Also, $p(x)$ has no $x^3$ term, so we insert $0 \cdot x^3$ as a placeholder — this prevents alignment errors during long division.

Divide the leading term of the dividend by the leading term of the divisor:

\frac{x^4}{x^2} = x^2

Multiply $x^2$ by the entire divisor $(x^2 - x + 1)$ :

x^2 \cdot (x^2 - x + 1) = x^4 - x^3 + x^2

Subtract from the dividend:

(x^4 + 0 \cdot x^3 - 3x^2 + 4x + 5) - (x^4 - x^3 + x^2) = x^3 - 4x^2 + 4x + 5

Bring down the next term (already done). Divide the new leading term:

\frac{x^3}{x^2} = x

Multiply $x$ by $(x^2 - x + 1)$ :

x \cdot (x^2 - x + 1) = x^3 - x^2 + x

Subtract:

(x^3 - 4x^2 + 4x + 5) - (x^3 - x^2 + x) = -3x^2 + 3x + 5

Divide the new leading term:

\frac{-3x^2}{x^2} = -3

Multiply $-3$ by $(x^2 - x + 1)$ :

-3 \cdot (x^2 - x + 1) = -3x^2 + 3x - 3

Subtract:

(-3x^2 + 3x + 5) - (-3x^2 + 3x - 3) = 8

The degree of the remainder (0) is less than the degree of the divisor (2), so we stop.

\text{Quotient: } q(x) = x^2 + x - 3

\text{Remainder: } r(x) = 8

Verification using Division Algorithm: $p(x) = g(x) \cdot q(x) + r(x)$

$(x^2 - x + 1)(x^2 + x - 3) + 8$

$= x^4 + x^3 - 3x^2 - x^3 - x^2 + 3x + x^2 + x - 3 + 8$

$= x^4 + (x^3 - x^3) + (-3x^2 - x^2 + x^2) + (3x + x) + (-3 + 8)$

$= x^4 - 3x^2 + 4x + 5$ ✓

Why This Works

Polynomial long division mirrors the long division you learned for integers. The key principle: at each step, we eliminate the highest degree term of the current remainder by choosing the right multiplier for the divisor.

The Division Algorithm for polynomials states: for any polynomial $p(x)$ and non-zero polynomial $g(x)$ , there exist unique polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that:

p(x) = g(x) \cdot q(x) + r(x)

where the degree of $r(x)$ is strictly less than the degree of $g(x)$ , or $r(x) = 0$ .

This is analogous to: dividend = divisor × quotient + remainder for integers.

Alternative Method

We can verify the remainder using the Remainder Theorem — but only when dividing by a linear factor $(x - a)$ . Since $g(x)$ is quadratic, the Remainder Theorem doesn’t directly apply here. Long division is the correct method for quadratic or higher-degree divisors.

For checking work quickly: substitute any convenient value of $x$ into both sides of $p(x) = g(x) \cdot q(x) + r(x)$ .

Try $x = 0$ : LHS = $p(0) = 5$ . RHS = $g(0) \cdot q(0) + 8 = 1 \cdot (-3) + 8 = 5$ . ✓

Try $x = 1$ : LHS = $1 - 3 + 4 + 5 = 7$ . RHS = $(1-1+1)(1+1-3) + 8 = 1 \times (-1) + 8 = 7$ . ✓

In CBSE Class 10 exams, polynomial division questions almost always ask you to verify using the division algorithm. Even if you’re confident your answer is right, write out the verification — it typically earns 1-2 marks on its own. The substitution check (trying $x = 0$ and $x = 1$ ) is a 30-second sanity check before writing the final answer.

Common Mistake

The most common error is forgetting to insert the missing $x^3$ term in $p(x)$ . If you write $p(x) = x^4 - 3x^2 + 4x + 5$ and jump into division without the placeholder, you’ll misalign terms in the subtraction step and get completely wrong coefficients in the quotient. Always write polynomial dividends with all terms from the highest degree down to the constant, inserting $0 \cdot x^k$ for any missing degree $k$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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