MAT1511
ASSIGNMENT 1
DUE DATE: MARCH 2026
, MAT1511 ASSIGNMENT 1 2026
DUE MARCH 2026
Question 1
Given: P(x) = 2x³ − 7x² + 4x + 4
(a) Possible Rational Zeros
Using the Rational Root Theorem:
Factors of constant term (4): ±1, ±2, ±4
Factors of leading coefficient (2): ±1, ±2
Possible rational zeros: ±1, ±2, ±4, ±1/2
(b) Descartes’ Rule of Signs
P(x) = 2x³ − 7x² + 4x + 4
Signs: +, −, +, + → 2 sign changes
Possible positive zeros: 2 or 0
P(−x) = −2x³ − 7x² − 4x + 4
Signs: −, −, −, + → 1 sign change
Possible negative zeros: 1
Total degree = 3 - Remaining roots are complex if not real.
ASSIGNMENT 1
DUE DATE: MARCH 2026
, MAT1511 ASSIGNMENT 1 2026
DUE MARCH 2026
Question 1
Given: P(x) = 2x³ − 7x² + 4x + 4
(a) Possible Rational Zeros
Using the Rational Root Theorem:
Factors of constant term (4): ±1, ±2, ±4
Factors of leading coefficient (2): ±1, ±2
Possible rational zeros: ±1, ±2, ±4, ±1/2
(b) Descartes’ Rule of Signs
P(x) = 2x³ − 7x² + 4x + 4
Signs: +, −, +, + → 2 sign changes
Possible positive zeros: 2 or 0
P(−x) = −2x³ − 7x² − 4x + 4
Signs: −, −, −, + → 1 sign change
Possible negative zeros: 1
Total degree = 3 - Remaining roots are complex if not real.
