Factors of Polynomials

• Carry this away. Divisor zero gives the test value; the test value gives the remainder; a zero remainder gives a factor; a factor gives a root. Five words in order: divisor, value, remainder, factor, root.
• Pause and recall. Cover the page. Say aloud: what is the remainder when \(P(x)\) is divided by \(x-a\)? Then: what is the test value for the divisor \(3x+2\)? Then: does a remainder of \(5\) tell you anything about factors? Answers: \(P(a)\); \(x=-\tfrac23\); yes — it tells you \(3x+2\) is not a factor.
• Make it a habit. Before you write your final line, run the two coefficient checks. They take about five seconds, they need no calculator, and between them they catch a dropped leading coefficient and a wrong sign — the two most expensive errors in this chapter.
• Scoring yourself honestly. Four out of four with full working and a check on each means this chapter is secure. Three or fewer means go back to the section named by the tag on the question you missed — not to the answer you just read.
• The one-minute version. If you only have a minute before an exam, recite: solve the divisor equal to zero; substitute; zero means factor; divide; factorise the quadratic; state every root. That sentence is the chapter.

• I can state the remainder theorem correctly: the remainder on dividing \(P(x)\) by \(x-a\) is \(P(a)\).
• I can find the test value for any linear divisor by solving the divisor equal to zero — including \(x+a\) and \(bx-c\).
• I can evaluate \(P(a)\) accurately for negative and fractional \(a\) without sign slips.
• I can state the factor theorem in both directions: \(x-a\) is a factor if and only if \(P(a)=0\).
• I can use a given factor or a given remainder to find an unknown coefficient in a polynomial.
• I can choose sensible candidate values to test, using the constant term and the leading coefficient, and I verify each one rather than assuming it.
• I can divide a cubic by a linear factor using polynomial long division, including inserting a zero placeholder for a missing power.
• I can carry out the same division by comparing coefficients when that is quicker.
• I can factorise the resulting quadratic and write the polynomial as a product of a linear factor and a quadratic factor, and where possible as a product of three linear factors.
• I can check my factorisation by expanding it back and by comparing the leading coefficient and the constant term.
• I can take a cubic equation from a discovered root all the way to a complete solution set, without stopping at the first root.
• I can solve the remaining quadratic by factorisation or by the quadratic formula, and I keep surd roots exact.
• I can recognise a repeated factor and describe the repeated root correctly.
• I can recognise when the quadratic factor has no real roots, so the cubic has exactly one real root.
• I can decide whether a root must be rejected, and I reject one only when the question states a restriction that forbids it.

Build your own questions — it is unusually easy here. Pick any three brackets, say \((x-4)(2x+3)(x+1)\), and expand them. You now have a cubic whose complete factorisation and roots you already know, so you can set yourself an unlimited supply of questions with guaranteed answers. Making them also drills the expansion that verifies every answer in this chapter.

Complete Factorisation

The expression of a polynomial as a product of factors that cannot be broken down further into brackets with integer coefficients. For a cubic this means either three linear factors, or one linear factor multiplied by a quadratic factor that does not factorise, because its discriminant is either negative or positive but not a perfect square. It is reached by using the factor theorem to find one linear factor, dividing it out to obtain a quadratic, and then factorising that quadratic if it factorises. A factorisation that stops at a linear factor times a quadratic that could still be factorised is incomplete.

Remainder Theorem

A result stating that when a polynomial P(x) is divided by a linear divisor x - a, the remainder is the single number P(a), obtained by substituting x = a into the polynomial. It converts a division problem into an evaluation problem, so the remainder can be found without carrying out the division. For a divisor written in another form, such as x + 2 or 2x - 1, the value substituted is the value that makes the divisor zero, namely -2 and one half respectively.

Polynomial Long Division

A column method for dividing one polynomial by another, laid out like numerical long division with the columns representing descending powers of x. At each stage the leading term of what remains is divided by the leading term of the divisor to give the next term of the quotient, that term is multiplied through the divisor, and the product is subtracted. Any power missing from the dividend must be written in with a zero coefficient so that the columns stay aligned. When the divisor is a factor, the process ends with a remainder of zero and the quotient is the cofactor.

Factor Theorem

A result stating that x - a is a factor of the polynomial P(x) if and only if P(a) = 0. It is the special case of the remainder theorem in which the remainder is zero, so the division leaves nothing over and P(x) can be written as x - a multiplied by another polynomial. The equivalence runs both ways: a zero value proves a factor, and a factor forces the value to be zero, so a non-zero value disproves the factor outright.

Root of a Polynomial

A value of x for which the polynomial P(x) evaluates to zero, and therefore a solution of the equation P(x) = 0. Each linear factor of the polynomial contributes one root, found by setting that factor equal to zero, so the factor x - a gives the root x = a and the factor 2x - 1 gives the root x = one half. A cubic with real coefficients has at most three real roots and at least one; a repeated linear factor produces a repeated root.

• “Dividing by \(x+2\), so I work out \(P(2)\).” Why it fails The theorem is stated for \(x-a\). Rewriting \(x+2\) as \(x-(-2)\) shows that \(a=-2\). Computing \(P(2)\) gives the remainder on division by \(x-2\), a different divisor, so the answer is not inaccurate — it answers another question entirely. Fix Solve \(x+2=0\) to get \(x=-2\), then compute \(P(-2)\).
• “Dividing by \(x-2\), so I work out \(P(-2)\).” Why it fails This is the same error running the other way, and it comes from half-recalling a “change the sign” rule. For \(P(x)=2x^3-5x^2+4x-7\) the two answers are \(-3\) and \(-51\); nothing about \(-51\) looks wrong, which is what makes the error costly. Fix Never change a sign. Solve the divisor equal to zero: \(x-2=0\) gives \(x=2\).
• “\(x+a\) is a factor, so \(x=a\) is a root.” Why it fails A root is the value that makes the factor zero, and \(x+a\) is zero when \(x=-a\). For \(P(x)=2x^3-3x^2-11x+6\), the factor \(x+2\) gives the root \(-2\), and indeed \(P(-2)=0\) while \(P(2)=-12\). Fix Read every factor by asking “what value makes this bracket zero?”
• “Roots are \(-2\), \(\tfrac12\) and \(3\), so the factors are \(x-2\), \(x+\tfrac12\) and \(x-3\).” Why it fails The sign flip has been applied in the wrong direction. The root \(-2\) corresponds to the factor \(x+2\), and the root \(\tfrac12\) to \(x-\tfrac12\), or equivalently \(2x-1\). Fix Root \(r\) gives factor \(x-r\). Substitute the actual number and simplify: \(r=-2\) gives \(x-(-2)=x+2\).
• “\(P(4)=2\), which is small, so \(x-4\) is approximately a factor.” Why it fails Factors are exact. \(P(4)=2\) means \(P(x)=(x-4)Q(x)+2\), and that \(+2\) is exactly what prevents \(x-4\) from dividing \(P(x)\). There is no “nearly divides” in polynomial division. Fix The condition is \(P(a)=0\), full stop. Report \(2\) as the remainder and move on to another candidate.
• “Dividing \(2x^3-5x^2+4x-7\) by \(x-2\) gives \(2x^2-x+2\), so the remainder is \(2x^2-x+2\).” Why it fails The quotient and the remainder are different objects. \(P(x)=(x-2)(2x^2-x+2)-3\): the quotient is the bracket that multiplies the divisor, and the remainder is the leftover constant \(-3\). Because the divisor is linear, the remainder must be a number — anything with an \(x\) in it cannot be the remainder. Fix If your “remainder” contains \(x\), you have named the quotient. The remainder is the number left at the foot of the division, and it equals \(P(a)\).
• “Divide \(x^3-7x+6\) by \(x-1\): bring down the terms as they are written.” Why it fails There is no \(x^2\) term, and long division is a column method whose columns are the powers of \(x\). Without an \(0x^2\) column, \(-7x\) is subtracted from the \(x^2\) column, and every subsequent term is one place out. The working still looks tidy, which is why the error survives to the final answer. Fix Write \(x^3+0x^2-7x+6\) before you start. Do it for every missing power, every time.
• “\(-3x^2\) minus \(-6x^2\) is \(-9x^2\).” Why it fails The two signs have been combined as though the operation were addition. Subtracting a negative adds: \((-3x^2)-(-6x^2)=-3x^2+6x^2=+3x^2\). One sign error here corrupts every later line, and long division gives you no warning — you simply reach a non-zero remainder for a divisor that really was a factor. Fix Change every sign in the row being subtracted, write the changed row down, and add. Do not hold two flips in your head.
• “\(P(3)=0\), so the solution of \(2x^3-3x^2-11x+6=0\) is \(x=3\).” Why it fails A cubic has up to three roots. \(x=3\) is one of them; \(x=\tfrac12\) and \(x=-2\) are equally valid and equally required. The word “solve” asks for the complete solution set. Fix A discovered root is the start of the question. Divide it out and finish the quadratic.
• “\(P(x)=(x-3)(2x^2+3x-2)\). Factorised.” Why it fails “Factorise completely” means no factor can be broken down further, and \(2x^2+3x-2=(2x-1)(x+2)\) plainly can be. The same answer would be complete if the quadratic did not factorise — if its discriminant were negative, or positive but not a perfect square — so the test is the discriminant, not the appearance of the expression. Fix Before declaring a factorisation complete, check the discriminant of the quadratic factor. If \(b^2-4ac\) is a perfect square, including \(0\), it factorises — keep going.
• “Solve \(P(x)=0\). Answer: \((x-3)(2x-1)(x+2)\).” Why it fails That is a factorisation, not a solution. A factorisation is an identity true for every \(x\); a solution is the particular set of values that make the expression zero. The work is complete but the question is unanswered. Fix Add the last line: set each bracket to zero and state \(x=-2,\ \tfrac12,\ 3\).
• “Factorise \(P(x)\) completely. Answer: \(x=-2,\ \tfrac12,\ 3\).” Why it fails The same confusion in reverse. Roots are numbers; a factorisation is a product of brackets. Note also that the roots alone do not determine the polynomial — \((x-3)(2x-1)(x+2)\) and \((x-3)(x-\tfrac12)(x+2)\) share those roots but are not the same polynomial. Fix Give the product of brackets, and check the leading coefficient reproduces the original \(2x^3\).
• “I got \((x-3)(2x^2+3x+2)\) from the division. Next question.” Why it fails Nothing checked it. Expanding takes fifteen seconds and would have shown \((x-3)(2x^2+3x+2)=2x^3-3x^2-7x-6\), which is not the original \(2x^3-3x^2-11x+6\). Even the two-second constant check catches it: \((-3)\times2=-6\), not \(+6\). Fix Always expand the final factorisation, or at minimum run the leading-coefficient and constant-term checks. A division error is invisible until you look.
• “The other roots are \(3.73\) and \(0.27\).” Why it fails The exact roots were \(2+\sqrt3\) and \(2-\sqrt3\). Rounding discards information the question never asked you to discard, and it makes the answer merely approximately true. On Paper 1 there is no calculator to produce those decimals with in the first place. Fix Leave surds as surds. Convert only when the question explicitly asks for a decimal or a degree of accuracy, and then only at the final step.

• Two papers, both compulsory, both two hours, both 80 marks, each worth 50% of the qualification, and both containing structured and unstructured questions. The difference that matters for this topic is the calculator rule.
• The consequence for this chapter. Paper 1 is a non-calculator paper, and this topic is built on substitution arithmetic — cubes of negative numbers, fractions such as \(P\!\left(\tfrac12\right)\), long chains of signed terms. All of that has to be done by hand, cleanly, under time pressure. So practise the arithmetic of outcome 3.1 without a calculator even when you are working through Paper 2 style questions. The method is never the hard part in this topic; the arithmetic is.
• You are given a polynomial and a linear divisor and asked for the remainder. One substitution, one number. The whole difficulty is getting the test value and the signs right.
• “Show that \(x-a\) is a factor of \(P(x)\)”. The substitution must be written out, the value \(0\) must appear, and the conclusion must be stated in words. A bare “\(=0\)” with no statement leaves the argument unfinished.
• A polynomial contains \(k\), or \(a\) and \(b\), and you are told a factor or a remainder. Each piece of information gives one equation. One unknown needs one condition; two unknowns need two, and you solve them simultaneously.
• The full chain, and the one that carries the most working: find a root, divide, factorise the quadratic, state the complete factorisation, then solve. When it is split into parts, one part asks for the factorisation and another for the roots, which is why answering the wrong one of those two costs marks in an otherwise perfect solution.