Factor Theorem

• State the Factor Theorem: (x − a) is a factor of p(x) iff p(a) = 0.
• Use factor theorem to test factors and find unknown coefficients.
• Factorise cubics by finding one zero, then dividing.
• Link zeros of polynomial to x-intercepts of graph (intro).
• Solve NCERT factorisation problems systematically.

1. Factor Theorem — (x − a) is a factor of p(x) if and only if p(a) = 0.
2. Converse of remainder — Remainder 0 ⇔ factor exists.
3. Finding factors — Try rational roots ±1, ±2, … for small integer coefficients.
4. Example — p(x) = x² − 5x + 6; p(2) = 0 and p(3) = 0 → factors (x − 2)(x − 3).
5. Cubic factorisation — If p(1) = 0, divide by (x − 1), factor the quadratic quotient.
6. Unknown coefficient — If (x − 1) is a factor of x² + kx + 3, then p(1) = 0 gives k.
7. (x + a) factor — Means p(−a) = 0.
8. NCERT Ex. 2.4 — Factorise x³ − 23x² + 142x − 120.
9. Multiple factors — (x − 1)² as repeated factor when p(1) = 0 and derivative also zero (extension).
10. Graph link — Zero at x = a ↔ graph crosses/c touches x-axis at a.

NCERT: Factorise x³ − 6x² + 11x − 6 using Factor Theorem

Step 1 — Try x = 1: p(1) = 1 − 6 + 11 − 6 = 0 → (x − 1) is a factor
Step 2 — Divide p(x) by (x − 1): quotient = x² − 5x + 6
Step 3 — Factor quadratic: x² − 5x + 6 = (x − 2)(x − 3)
Step 4 — p(x) = (x − 1)(x − 2)(x − 3)
Verify: expand product matches original ✓

• Testing p(−a) for factor (x − a).
• Stopping after one factor without fully factorising the quotient.
• Assuming any p(a)=0 means (x+a) is factor (sign of factor matters).
• Arithmetic errors when evaluating p(a) for cubics.
• Ignoring constant term when guessing trial roots (must divide constant term).

• Is (x − 3) a factor of x² − 8x + 15? Verify.
• Find k if (x − 2) is a factor of x² + kx − 8.
• One factor of x³ − 4x² + x + 6 is (x + 1). Find others.
• Write zeros of (x − 1)(x + 4).
• Factor Theorem in one sentence?

Open the Practice tab for graded questions on Factor Theorem.