Three linear factors
- Input
- x^3 - 6x^2 + 11x - 6
- Expected output
- (x − 1)(x − 2)(x − 3)
Integer roots 1, 2 and 3.
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factor a polynomial
Paste the polynomial on the Roots and factoring tab. The factorization over ℚ appears ready, with linear factors and irreducible quadratics, and the factorization over ℂ lists one root per factor.
Integer roots 1, 2 and 3.
Equivalent to (x + 2)(2x − 1)(x − 3).
1 is a double root.
Over ℂ, x² + 1 = (x − i)(x + i).
Roots 1 ± i and −1 ± i.
No rational root and of degree 3, hence irreducible over ℚ.
Divide the leading term of the dividend by the leading term of the divisor, multiply the result by the whole divisor, subtract it from the dividend and repeat with the running remainder until the remainder has a lower degree than the divisor. Example: (2x³ − 3x + 1) ÷ (x² + x − 2). First 2x³ ÷ x² = 2x; 2x·(x² + x − 2) = 2x³ + 2x² − 4x; what is left is −2x² + x + 1. Then −2x² ÷ x² = −2; −2·(x² + x − 2) = −2x² − 2x + 4; what is left is 3x − 3. Quotient 2x − 2, remainder 3x − 3. Missing terms, like the x² in the dividend, go in with a zero coefficient.
Over ℚ the factor coefficients are rational; over ℝ they are real, which allows factors like x − √2; over ℂ, every polynomial becomes a product of degree 1 factors. The tool shows ℚ and ℂ. For x² − 2, ℚ returns x² − 2 itself, and ℂ gives (x + √2)(x − √2).
Look for rational roots first: each one removes a linear factor. If degree 2 is left, apply the quadratic formula. If degree 4 is left with no rational root, like x⁴ + 4, try a product of two quadratics; the tool runs that search with Kronecker’s method.
Because the root is fractional: 1/2 is a root of 2x − 1, and x − 1/2 is the matching monic factor. Multiplying that factor by 2 and dividing the leading coefficient by 2 gives the equivalent form with integers only.
= 2x³ − 3x + 1
= x² + x − 2
Long division
Division stops when the degree of the remainder is less than the degree of the divisor.
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