Maths › Algebra and functions › Polynomials and the factor theorem
Polynomials and the factor theorem
Cubics and beyond stop being scary the moment you learn to interrogate them. One substitution tells you whether a bracket divides a polynomial exactly, long division does the splitting, and suddenly every well-behaved cubic is three brackets in a trench coat.
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 cubics completely.
- Simplify rational expressions by factorise-and-cancel and by algebraic division.
WHAT YOU PROBABLY THINK
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, expanding, collecting, 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, just as 17 ÷ 5 gives quotient 3, 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 each column has a place.
Divide leading terms in turn. 4x3 ÷ 2x = 2x2; multiply back, subtract, and repeat to get −2x, then −1, with 4 left over.
So the quotient is 2x2 − 2x − 1 and the remainder is 4, and the check is one expansion. (2x − 1)(2x2 − 2x − 1) + 4 = 4x3 − 6x2 + 5.
The missing 0x placeholder is not decoration. Skip it and every later subtraction lands in the wrong column.
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 doing the division.
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 just f(a), and in particular
which is the factor theorem. Mind the sign, because it retires the trap this lesson opened with. The bracket (x − 2) is zero when x = 2, so testing it means evaluating f(2), not f(−2). A divisor like (2x − 1) works the same way; substitute the value that kills the bracket, x = 1/2.
WORKED EXAMPLE
A cubic with a hidden square
Factorise completely f(x) = x3 + 3x2 − 4.
Hunt for a root among small integers. f(1) = 1 + 3 − 4 = 0, so (x − 1) is a factor.
Divide. x3 + 3x2 − 4 = (x − 1)(x2 + 4x + 4), and the quadratic is a perfect square.
So f(x) = (x − 1)(x + 2)2, one 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 to ±1, ±2 and ±4.
YOUR TURN
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 named as the factor theorem.
Rational expressions
An algebraic fraction simplifies the same 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 that later chapters on curve sketching and integration lean on heavily.
TRY IT UNSEEN
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 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 become a stopping rule. A cubic always has at least one real root, but it is never obliged to have three.
THE EXAM BIT
- Write dividends with every power present, inserting 0x terms, before dividing. 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, not the answer.
- For a factor (ax − b), substitute x = b/a, the value that makes the bracket zero. The sign and the fraction are both tested deliberately.
- Candidate integer roots divide the constant term. Say so, then test them in order; markers like seeing the shortlist.
- Cancel whole factors only, never terms, and quote the identity f(x) = divisor × quotient + remainder when asked to interpret a division.
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 the quadratic factorises.
f(x) = (x + 1)(x − 3)(x + 2), three simple roots, and expanding back confirms it in one line.
The remainder on dividing f(x) by (x − a) is f(a); a zero remainder names a factor.
Factorise cubics by one found root, one division, and a quadratic you already know how to finish.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.
- 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 cubics completely.
- Simplify rational expressions by factorise-and-cancel and by algebraic division.
No animated video for this topic yet; these notes stand alone.