IBDP Maths AHL 2.12 Polynomial functions and their graphs AA HL Paper 1- Exam Style Questions- New Syllabus

Method 1 – Using the double‑root condition
Let \( f(x) = x^4 + px^3 – 2x^2 + qx – 3 \).
Since \( (x+1)^2 \) is a factor, \( x = -1 \) is a repeated root. This implies that \( f(-1) = 0 \) and its derivative \( f'(-1) = 0 \).
1. \( f(-1) = (-1)^4 + p(-1)^3 – 2(-1)^2 + q(-1) – 3 = 0 \)
\( 1 – p – 2 – q – 3 = 0 \implies p + q = -4 \). (Equation 1)
2. Differentiating \( f(x) \): \( f'(x) = 4x^3 + 3px^2 – 4x + q \).
\( f'(-1) = 4(-1)^3 + 3p(-1)^2 – 4(-1) + q = 0 \)
\( -4 + 3p + 4 + q = 0 \implies 3p + q = 0 \). (Equation 2)
Subtracting Equation 1 from Equation 2:
\( (3p + q) – (p + q) = 0 – (-4) \implies 2p = 4 \implies p = 2 \).
Substituting \( p = 2 \) into Equation 1:
\( 2 + q = -4 \implies q = -6 \).
\( \boxed{p = 2,\; q = -6} \)

Method 2 – Comparing Coefficients
Since \( (x+1)^2 = x^2 + 2x + 1 \) is a factor, we can write:
\( x^4 + px^3 – 2x^2 + qx – 3 = (x^2 + 2x + 1)(x^2 + ax – 3) \)
(Note: the constant term must be \(-3\) to satisfy \( 1 \times -3 = -3 \)).
Expanding the right side:
\( x^4 + ax^3 – 3x^2 + 2x^3 + 2ax^2 – 6x + x^2 + ax – 3 \)
Group like terms:
\( x^4 + (a+2)x^3 + (2a-2)x^2 + (a-6)x – 3 \).
Comparing the coefficient of \( x^2 \):
\( 2a – 2 = -2 \implies 2a = 0 \implies a = 0 \).
Now compare coefficients of \( x^3 \) and \( x \):
\( p = a + 2 = 0 + 2 = 2 \).
\( q = a – 6 = 0 – 6 = -6 \).
\( \boxed{p = 2,\; q = -6} \)