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.

• “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 \(0.268\).” 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. Note that the decimals here are correctly rounded to three significant figures and the answer is still wrong — the fault is converting at all, not converting badly. 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.
• Both papers, every candidate. Both papers are compulsory and every candidate takes both — 0606 has no Core/Extended tier split, so there is no easier paper and no harder paper to be entered for. Either paper may set a question on any part of the syllabus, so Topic 3 belongs to neither one exclusively. Both papers carry structured and unstructured questions, both require you to show all necessary working, and both assess AO1 (knowledge and understanding of mathematical techniques) and AO2 (analyse, interpret and communicate mathematically) at roughly 45–55% each. Grades A* to E are available; F and G are not.
• The examination runs in the June and November series, and in the March series in India. The course assumes you have covered Cambridge IGCSE Mathematics or an equivalent, which is where the quadratic work this chapter leans on comes from.
• 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.