Chapter 2.5 — Exercise 2.4 — Factor Theorem

The Two Directions of the Theorem

The "if and only if" here genuinely runs both ways, and the exercise uses both directions. If p(a) = 0, then p(x) = (x − a)·q(x) for some polynomial q(x), which is exactly what "factor" means. Conversely, if (x − a) really is a factor, then p(x) = (x − a)·q(x), and substituting x = a immediately gives p(a) = (a − a)·q(a) = 0. Neither direction requires knowing q(x) explicitly — the whole theorem operates purely on the value p(a). That's what makes it faster than actually carrying out a division to check for a factor: the Factor Theorem answers a yes/no question about factors using nothing more than the same substitute-and-simplify arithmetic already familiar from finding zeroes and remainders earlier in the chapter.

Two Coefficient-Sum Shortcuts

Two special divisors come up often enough to deserve their own rule. Testing whether (x − 1) is a factor means checking p(1) — which just adds up every coefficient in the polynomial, since 1 raised to any power is still 1. So (x − 1) is a factor exactly when all the coefficients sum to zero. Testing (x + 1) means checking p(−1), where alternating signs separate terms by whether their power is even or odd — so (x + 1) is a factor exactly when the even-power coefficients sum to the same value as the odd-power coefficients.

Applied to x³ − x² − x + 1: even-power coefficients (the x² term and the constant) sum to −1 + 1 = 0; odd-power coefficients (x³ and x) sum to 1 − 1 = 0. Equal sums, so (x + 1) is a factor — confirmed directly by p(−1) = −1 − 1 + 1 + 1 = 0. Run the same check on x⁴ − x³ + x² − x + 1: even-power terms sum to 1 + 1 + 1 = 3, odd-power terms sum to −1 − 1 = −2. Unequal, so (x + 1) is not a factor here. Both shortcuts save real work compared to the general method: instead of finding a linear divisor's zero and substituting a potentially awkward fraction, checking (x−1) or (x+1) reduces to nothing more than adding a short list of whole-number coefficients — arithmetic simple enough to do by inspection, without writing out a single substitution.

Confirming a Factor for Any Linear Divisor

Beyond the ±1 shortcuts, the same p(a) = 0 check works for any linear divisor once its zero is identified. For f(x) = 3x³ + x² − 20x + 12 and g(x) = 3x − 2, the zero of g is x = 2/3, and f(2/3) = 3(8/27) + 4/9 − 40/3 + 12 = 8/9 + 4/9 − 40/3 + 12 = 0 — confirming g(x) is a factor of f(x). The arithmetic gets heavier as the divisor's zero becomes a less convenient fraction, but the check itself never changes shape. It's worth noticing that a fractional zero like 2/3 isn't a warning sign of a mistake somewhere — many genuinely factorable cubics have divisors whose zero simplifies to something other than a whole number, and the substitution still resolves to exactly 0 when the factor really does divide evenly.

Confirming Three Factors at Once

Some problems give a cubic and three candidate linear factors together, asking for all three to be verified in one pass. For x³ − 3x² − 10x + 24: f(2) = 8 − 12 − 20 + 24 = 0, f(−3) = −27 − 27 + 54 = 0, and f(4) = 64 − 48 − 16 = 0 — three separate substitutions, each one independently confirming (x − 2), (x + 3), and (x − 4) as factors. Since a cubic has at most three linear factors, finding three that all check out means the complete factorisation has been found without any further work: x³ − 3x² − 10x + 24 = (x − 2)(x + 3)(x − 4). A related problem gives x³ − 6x² − 19x + 84 with candidates (x + 4), (x − 3), and (x − 7), each verified the same way — f(−4) = −64 − 96 + 160 = 0, f(3) = 27 − 54 + 27 = 0, f(7) = 343 − 294 − 49 = 0 — landing on the same style of result: three roots confirmed independently, no long division required at all.

Factorising a Cubic Completely

The exercise's main task combines factor-checking with division: find one factor by testing small values, divide it out, then factorise whatever quadratic remains. For x³ − 2x² − x + 2: f(1) = 1 − 2 − 1 + 2 = 0 and f(−1) = −1 − 2 + 1 + 2 = 0, so both (x − 1) and (x + 1) — together, (x² − 1) — are factors. Dividing x³ − 2x² − x + 2 by x² − 1 gives quotient (x − 2) exactly, so the full factorisation is (x − 1)(x + 1)(x − 2).

A second case, x³ − 3x² − 9x − 5, only has one linear factor findable by direct testing: f(−1) = −1 − 3 + 9 − 5 = 0. Dividing by (x + 1) leaves quotient x² − 4x − 5, which splits by the middle term as x² − 5x + x − 5 = (x − 5)(x + 1) — so the complete factorisation is (x + 1)²(x − 5), a repeated factor rather than three distinct ones.

Shared Factors Between Two Polynomials

The harder problems involve two polynomials with a factor in common, translated into simultaneous equations. If both (x − 2) and (x − ½) are factors of px² + 5x + r, then f(2) = 0 and f(½) = 0 give 4p + 10 + r = 0 and p + 10 + 4r = 0 — subtracting these two equations cleanly eliminates the constant 10, leaving 3p = 3r, so p = r. A similar setup — x² − x − 6 and x² + 3x − 18 sharing a common factor (x − a) — turns into a² − a − 6 = 0 and a² + 3a − 18 = 0 simultaneously, which combine to give 4a = 12, so a = 3.

Into Algebraic Identities

Every technique in this exercise — the coefficient-sum shortcuts, dividing out a confirmed factor, solving simultaneous factor conditions — depends on being able to substitute quickly and reliably, the skill built across the whole chapter so far. Exercise 2.5 takes a different route to factorisation entirely, recognising standard algebraic patterns instead of testing candidate zeroes one at a time — a genuinely different strategy for the same underlying goal.