Polynomial Division and the Factor Theorem
Divide a polynomial by (x - k), find its remainder, and use a root to split a cubic into a linear factor and a quadratic quotient.
What a learner can do afterwards
- Divide a cubic by (x - k), state the quotient and remainder, and check by multiplication
- Use f(k) = 0 to identify the factor (x - k), including when k is negative
- Find integer-root candidates, divide out a confirmed factor, and factorise the quadratic quotient when possible
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Last stop you checked allowed inputs for the Ridgeway pump rule. Here, an output of zero gives a polynomial factor. Find a root of a cubic, then divide out its factor to leave a quadratic. With whole-number coefficients and a nonzero constant term, any integer root must divide that constant term. Try both signs of its divisors. These are candidates, not confirmed roots. Some cubics need another method, and some quadratic quotients have no further integer factors.
For a polynomial, division by (x - k) gives f(x) = (x - k)Q(x) + R. Q(x) is the quotient; R is a constant remainder. Substitute x = k: the product is zero, so f(k) = R. This is the remainder theorem. Thus f(k) = 0 exactly when x - k is a factor, which is the factor theorem. It works in both directions. For k = -4, the matching factor is x + 4, since -4 + 4 = 0.
Write terms in descending powers. Insert 0 for missing powers, as in x³ + 0x² - 4x + 1. Divide the current leading term by the divisor's leading term. Multiply that result by the whole divisor, subtract the whole product, then bring down the next term. Repeat until only a constant remains. This remainder may be negative. A cubic divided by a linear expression has a quadratic quotient: x³ / x = x², and 2x³ / x = 2x².
Divide x³ - 2x² - 5x + 6 by x - 1. First, x³ / x = x². Multiply: x²(x - 1) = x³ - x². Subtract and bring down -5x to get -x² - 5x. Next, -x² / x = -x. Multiply: -x(x - 1) = -x² + x. Subtracting gives -6x. Bring down 6 to get -6x + 6. Each subtraction removes the current leading term.
For f(x) = x³ - 2x² - 5x + 6, f(1) = 1 - 2 - 5 + 6 = 0, so x - 1 is a factor. Finish the division: -6(x - 1) = -6x + 6, leaving remainder 0 and quotient x² - x - 6. Find two numbers with product -6 and sum -1: -3 and 2. Thus f(x) = (x - 1)(x - 3)(x + 2). Set each bracket to zero to get roots 1, 3 and -2. After dividing out a confirmed factor, factorise the quotient if possible.
Let p(x) = x³ + 2x² - 3x + 4. On division by x - 2, the remainder is p(2) = 8 + 8 - 6 + 4 = 14. Long division gives quotient x² + 4x + 5. Therefore p(x) = (x - 2)(x² + 4x + 5) + 14. Since the remainder is not zero, x - 2 is not a factor. Substitution finds the remainder; division also finds the quotient.
To evaluate a polynomial, replace every x with the input. Calculate powers, then products, then sums. Put negative inputs in parentheses. An unknown coefficient stays as a letter. For g(x) = x³ + ax - 12, g(2) = 8 + 2a - 12 = 2a - 4. If 2 is a root, set 2a - 4 = 0, giving a = 2. If division by x - 2 has remainder 6, set 2a - 4 = 6 instead, giving a = 5.
Check a division by multiplying the divisor by the quotient and adding the remainder. Expand and compare every coefficient with the dividend. This can find a missing remainder. For example, (x - 1)(x² + x + 2) = x³ + x - 2. To obtain x³ + x + 5, add R with -2 + R = 5, so R = 7. A factorisation has remainder 0, so multiplying its factors must give the original polynomial.
A repeated factor gives a repeated root, not another different root. For example, 2(x - 1)²(x + 3) has two distinct roots: 1 and -3. Keep the repeated factor in the full factorisation. Also keep the leading coefficient: this product starts with 2x³. Knowing the roots does not tell you that the leading coefficient is 1. A cubic need not have three different real roots.
Dividing by x - k leaves remainder f(k). If f(k) = 0, x - k is a factor. Divide it out, factorise the quotient if possible, and check by multiplying back.
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More to read
Following the powers through division
Polynomial division works one leading term at a time. Put the terms in descending powers and leave a place for every power. A missing term has coefficient 0.
For example, divide x³ + 3x² + 2x + 6 by x + 3. The first quotient term is x². Multiplying gives x³ + 3x², so subtraction removes both terms. The remaining expression is 2x + 6. There is no x term in the quotient because no x² term remains to divide. The next quotient term is 2, and 2(x + 3) removes the rest.
The quotient is x² + 2 and the remainder is 0. Check by expanding (x + 3)(x² + 2). For a division with a remainder, add that remainder after multiplying. The result must recover the dividend.
From a remainder to a factor
Suppose division by x - k gives quotient Q(x) and remainder R. Then f(x) = (x - k)Q(x) + R. At x = k, the product disappears and f(k) = R. This explains why substitution can find the remainder without finding the quotient.
If f(k) is zero, the division is exact and x - k is a factor. The reverse also holds: a factor x - k gives f(k) = 0. For the divisor x + k, use the input -k, because that is where its bracket is zero.
For f(x) = x³ + 2x - 3, f(1) = 1 + 2 - 3 = 0. Dividing out x - 1 leaves x² + x + 3. No pair of integers has product 3 and sum 1, so the integer-bracket method stops there. Finding one factor does not promise three integer linear factors.
Checking every part of a factorisation
A proposed factorisation must reproduce the whole polynomial. Checking roots can help you find brackets, but the leading coefficient still matters. Three brackets beginning with x have product beginning with x³. A polynomial beginning with 3x³ needs an overall factor of 3, or an equivalent coefficient inside one bracket.
Repeated brackets also matter. The expression 3(x + 2)²(x - 5) has the root -2 twice and the root 5 once. It has two different root values, not three. Keep both copies of x + 2 in its factorisation.
For a division check, multiply the divisor and quotient and add the remainder. Compare like powers, including the constant term. Subtracting a product means subtracting all of its terms. A negative remainder is allowed.
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