Compute \(Q(x)\) and \(R(x)\) such that \(P(x)=D(x)\,Q(x)+R(x)\) with \(\deg R<\deg D\).
\( \text{Euclidean division: } P(x)=D(x)\,Q(x)+R(x),\; \deg R<\deg D. \)
\( \text{Only two inputs are required: dividend and divisor (variable defaults to }x\text{).} \)
\( \text{If } \deg P < \deg D,\; Q(x)=0,\; R(x)=P(x). \)
A Dividing Polynomials Calculator divides one polynomial (the dividend) by another (the divisor) to produce a quotient and a remainder. It supports classical long division for divisors of any degree and synthetic division for linear divisors of the form \(x-a\). By inserting zeros for missing powers and aligning like terms, the calculator systematically cancels the dividend’s leading term at each step, building the quotient term-by-term. The output includes the quotient \(Q(x)\), remainder \(R(x)\), and a verification that \(P(x)=D(x)Q(x)+R(x)\). This is useful for simplifying rational expressions, finding oblique asymptotes, evaluating with the Remainder Theorem, and checking factorization work.
For polynomials \(P(x)\) and \(D(x)\) with \(\deg P\ge \deg D\) and nonzero leading coefficients, there exist unique \(Q(x)\) and \(R(x)\) such that
\[ P(x)=D(x)\,Q(x)+R(x),\qquad \deg R<\deg D. \]
Long division cancels the current remainder’s leading term by multiplying the divisor with a suitable monomial and subtracting:
\[ k(x)=\frac{\operatorname{lc}(P)}{\operatorname{lc}(D)}\,x^{\deg P-\deg D},\qquad P\leftarrow P-k(x)D(x). \]
Synthetic division accelerates division by \(x-a\) using a simple recurrence on coefficients, directly yielding the remainder \(R=P(a)\) (Remainder Theorem). The calculator normalizes sign conventions, supports exact arithmetic for integer/fractional coefficients, and flags undefined operations (e.g., zero divisor).
Division algorithm: \[ P(x)=D(x)\,Q(x)+R(x),\quad \deg R<\deg D. \]
Leading-term step (long division): \[ k(x)=\frac{\mathrm{lc}(P)}{\mathrm{lc}(D)}\,x^{\deg P-\deg D},\quad P\leftarrow P-k(x)D(x). \]
Synthetic division for \(x-a\): If \(P(x)=\sum_{j=0}^{n} c_j x^{\,n-j}\), \[ b_0=c_0,\quad b_k=c_k+a\,b_{k-1}\ (k=1,\dots,n),\quad Q(x)=\sum_{j=0}^{n-1} b_j x^{\,n-1-j},\ R=b_n=P(a). \]
Remainder Theorem: \[ \text{If } D(x)=x-a,\ \text{then } R=P(a). \]
Divide \(P(x)=2x^3+3x^2-11x+6\) by \(x-2\). Coefficients: \(2,3,-11,6\); use \(a=2\). Recurrence: \(b_0=2\); \(b_1=3+2\cdot2=7\); \(b_2=-11+2\cdot7=3\); \(b_3=6+2\cdot3=12\). Quotient \(Q(x)=2x^2+7x+3\), remainder \(R=12\). Check: \((x-2)Q(x)+12=P(x)\).
Divide \(P(x)=3x^4 - x^3 + 2x - 5\) (note the missing \(x^2\) term) by \(D(x)=x^2-1\). Steps yield \(Q(x)=3x^2 - x + 3\) and \(R(x)=x-2\). Verify: \(P=DQ+R\).
\(P(x)=x^3-1\) by \(x-1\): remainder \(P(1)=0\Rightarrow\) exact division with \(Q(x)=x^2+x+1\).
Long division works for any divisor; synthetic division is a faster shortcut only for divisors \(x-a\).
They keep powers aligned so additions and subtractions occur on like terms during division.
If the remainder \(R(x)=0\), then \(D(x)\) divides \(P(x)\) exactly.
Yes—use long division; the quotient will have degree \(\deg P-\deg D\) (or lower), with a remainder of degree less than \(\deg D\).
For \(x-a\), the remainder equals \(P(a)\). Nonzero remainders quantify how far from exact divisibility you are.
Yes. Always write polynomials in descending powers; otherwise steps and alignment will be incorrect.