2.1 Use Descartes' Rule of Signs to describe all possibilities for the number of positive, negative and imaginary zeros of \( P(x) \), where \[ P(x)=4 x^{3}-3 x^{2}-7 x+9 \] (You may summarise your answer in the form of a table.)

We begin by applying Descartes’ Rule of Signs to P(x):  P(x) = 4x³ – 3x² – 7x + 9 Step 1. Count sign changes for positive zeros: • The coefficients are: +4, –3, –7, +9.  - Transition from +4 to –3 is one sign change.  - Transition from –3 to –7 has no sign change (both are negative).  - Transition from –7 to +9 is a second sign change. • So there are 2 sign changes. Thus, the number of positive real zeros is either 2 or 0 (2, minus an even number). Step 2. Count sign changes for negative zeros: • Substitute x = –y to obtain P(–x):  P(–x) = 4(–x)³ – 3(–x)² – 7(–x) + 9       = –4x³ – 3x² + 7x + 9 • The coefficients then are: –4, –3, +7, +9.  - –4 to –3: no sign change (both negative).  - –3 to +7: one sign change.  - +7 to +9: no sign change (both positive). • So there is 1 sign change. Hence, the number of negative real zeros is exactly 1. Step 3. Determine the number of imaginary (non‐real complex) zeros: • Since P(x) is a polynomial of degree 3, there must be exactly 3 zeros (counted with multiplicity). • Two cases arise:  Case 1: If there are 2 positive and 1 negative real zero, then there are 0 non‐real (imaginary) zeros.  Case 2: If there are 0 positive and 1 negative real zero, then the other 2 (complex conjugate) zeros must be non‐real. Summarized in a table: -----------------------------------------------------         | Positive | Negative | Imaginary ----------------------------------------------------- Case 1   | 2 | 1 | 0 Case 2   | 0 | 1 | 2 ----------------------------------------------------- Thus, by Descartes’ Rule of Signs for P(x) = 4x³ – 3x² – 7x + 9, the zeros can either be 2 positive, 1 negative, and 0 imaginary or 0 positive, 1 negative, and 2 imaginary (which, by the Conjugate Root Theorem, are a complex conjugate pair).