TMUA Polynomials

Polynomials on TMUA Paper 1 are almost always set up so that one root or one factor is gettable by inspection, and the rest of the work is what you do with that. The test rewards candidates who can step from a partial fact to the full factorisation cleanly.

• Factor theorem. If $p(a) = 0$ then $(x - a)$ is a factor of $p$, and vice versa. Integer roots of an integer-coefficient polynomial divide the constant term; test those first.
• Remainder theorem. $p$ divided by $(x - a)$ leaves remainder $p(a)$. Used to pin down one parameter when a remainder is given.
• Polynomial long division. Once one factor is in hand, divide it out to drop the degree by one. The quotient is usually a quadratic you can solve directly.
• Vieta’s relations for quadratics. For $ax^2 + bx + c$ with roots $\alpha, \beta$: $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. TMUA uses these to find expressions like $\alpha^2 + \beta^2$ or $\tfrac{1}{\alpha} + \tfrac{1}{\beta}$ without solving for the roots explicitly.

The move. When a TMUA polynomial question gives you a value of $p$ at some point, that’s the factor theorem in disguise. When it gives you a relation between roots, that’s Vieta in disguise. Identify which one the question is hiding, then the algebra is short.