Maths › Algebra and functions › Polynomials and the factor theorem
Polynomials and the factor theorem
Cubics stop being frightening the moment you learn to interrogate them. One substitution tells you whether a bracket divides a polynomial exactly, division does the splitting, and a well-behaved cubic turns out to be three brackets multiplied together.
Builds on Quadratic functions.
IN THIS TOPIC
- Divide a polynomial by a linear expression of the form (ax ± b), quotient and remainder included.
- Use the factor theorem to find factors, fix unknown coefficients, and factorise a cubic completely.
- Simplify rational expressions by factorise-and-cancel and by algebraic division.
COMMON MISCONCEPTION
To test whether (x − 2) is a factor, evaluate f(−2).
Dividing polynomials
A polynomial is a sum of whole-number powers of x, and the familiar moves of expanding, collecting and factorising all still apply at degree three and above. The new tool is algebraic division. Dividing f(x) by a linear expression produces a quotient and a remainder, tied together by the identity f(x) = divisor × quotient + remainder. Whole numbers work the same way. 17 ÷ 5 gives quotient 3 and remainder 2, because 17 = 5 × 3 + 2.
WORKED EXAMPLE
Division by (ax − b), remainder and all
Divide 4x3 − 6x2 + 5 by (2x − 1).
Write the dividend with every power present, 4x3 − 6x2 + 0x + 5, so that each column has something in it.
Divide leading terms in turn. 4x3 ÷ 2x = 2x2; multiply back, subtract, and repeat to get −2x, then −1, with 4 left over.
The quotient is 2x2 − 2x − 1 and the remainder is 4. One expansion checks it, since (2x − 1)(2x2 − 2x − 1) + 4 = 4x3 − 6x2 + 5.
That 0x placeholder is not decoration. Skip it and every later subtraction lands in the wrong column, which is the single most common way this topic goes wrong.
When the remainder comes out as zero, the divisor is a factor and the division has split the polynomial into a product. The next result predicts that outcome without making you do the division at all.
The factor theorem
Substituting x = a into f(x) = (x − a) × quotient + remainder makes the bracket vanish and leaves f(a) = remainder. So the remainder on dividing by (x − a) is simply f(a), and in particular
which is the factor theorem. The booklet does not print it, so the statement has to come from you. Mind the sign. The bracket (x − 2) is zero at x = 2, so testing it means evaluating f(2), and the sign trap catches people every summer. A divisor like (2x − 1) works the same way. Substitute the value that makes the bracket zero, here x = 1/2.
WORKED EXAMPLE
A cubic with a hidden square
Factorise completely f(x) = x3 + 3x2 − 4.
Hunt for a root among the small integers. f(1) = 1 + 3 − 4 = 0, so (x − 1) is a factor.
Divide. x3 + 3x2 − 4 = (x − 1)(x2 + 4x + 4), and that quadratic is a perfect square.
So f(x) = (x − 1)(x + 2)2, a simple root at 1 and a repeated root at −2.
Sensible candidates divide the constant term. Any integer root of this cubic must divide 4, which cuts the hunt down to ±1, ±2 and ±4.
GUIDED PRACTICE
A leading coefficient joins in
Show that (2x − 1) is a factor of f(x) = 2x3 − x2 − 8x + 4, and factorise f(x) completely, before opening the working.
Show the working
The bracket is zero at x = 1/2, and f(1/2) = 1/4 − 1/4 − 4 + 4 = 0, so (2x − 1) is a factor.
Divide, or spot the grouping 2x3 − x2 − 8x + 4 = x2(2x − 1) − 4(2x − 1) = (2x − 1)(x2 − 4).
The difference of two squares finishes it. f(x) = (2x − 1)(x − 2)(x + 2).
A show-that question wants the substitution shown, the zero stated, and the conclusion attributed to the factor theorem by name.
Rational expressions
An algebraic fraction simplifies the way a numerical one does. Factorise top and bottom, then cancel any factor they share. For (x2 − x − 6)/(x2 − 9), factorising gives (x − 3)(x + 2) over (x − 3)(x + 3), and cancelling the shared bracket leaves (x + 2)/(x + 3). Only whole factors cancel. Crossing out the x2 terms, or the 6 against the 9, is the classic wreck.
When the numerator's degree reaches the denominator's, the fraction is improper, and algebraic division rewrites it as a polynomial plus a proper remainder term. Dividing gives (2x2 + 3x − 4)/(x − 1) = 2x + 5 + 1/(x − 1), a form the later chapters on curve sketching and integration lean on heavily.
INDEPENDENT PRACTICE
An unknown coefficient
(x + 3) is a factor of f(x) = x3 + ax2 − x + 6. Find a, and factorise f(x) as far as possible with real factors.
Show the working
The factor theorem at x = −3 gives −27 + 9a + 3 + 6 = 0, so 9a = 18 and a = 2.
Divide by (x + 3). x3 + 2x2 − x + 6 = (x + 3)(x2 − x + 2).
Check the quadratic before hunting any further. Its discriminant is 1 − 8 = −7, negative, so it has no real roots and (x + 3)(x2 − x + 2) is as far as real factorisation goes.
Last lesson's discriminant has quietly become a stopping rule. Every cubic has at least one real root, but it is under no obligation to have three.
ASSESSMENT FOCUS
- Write dividends with every power present, inserting 0x terms, before you divide. Column slips in division are the most common lost method mark in this topic.
- State the factor theorem check explicitly. The substitution, the zero, the conclusion. “Show that” questions mark the reasoning and give nothing for the answer alone.
- For a factor (ax − b), substitute x = b/a, the value that makes the bracket zero. Both the sign and the fraction are being tested deliberately.
- Candidate integer roots divide the constant term. Write the shortlist out, then test it in order; markers like seeing that you knew where to look.
- Cancel whole factors only, never terms. When asked to interpret a division, quote f(x) = divisor × quotient + remainder and point at the piece the question wants.
CHECK YOURSELF
Show that (x + 1) is a factor of f(x) = x3 − 7x − 6, and hence factorise f(x) completely.
Show a hint
f(−1) first; then divide and factorise the quadratic quotient.
Show the answer
f(−1) = −1 + 7 − 6 = 0, so by the factor theorem (x + 1) is a factor.
Dividing gives x3 − 7x − 6 = (x + 1)(x2 − x − 6), and that quadratic factorises.
f(x) = (x + 1)(x − 3)(x + 2), three simple roots, and expanding back confirms it in one line.
Divide f(x) by (x − a) and the remainder is f(a). A remainder of zero names a factor.
One found root, one division, one quadratic you already know how to finish.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the polynomials and the factor theorem questions page.
CHECK YOUR PROGRESS
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- Divide a polynomial by a linear expression of the form (ax ± b), quotient and remainder included.
- Use the factor theorem to find factors, fix unknown coefficients, and factorise a cubic completely.
- Simplify rational expressions by factorise-and-cancel and by algebraic division.
Open the full revision checklist to see every objective in the course in one place.