MAT1581 Assignment 1 2026 Due 15 May 2026 |Mathematics I (Engineering)|

UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology



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MAT1581
ASSIGNMENT 01

Binomial Theorem · Determinants · Cramer’s Rule · Partial Fractions



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Module Code: MAT1581

Module Name: Mathematics I (Engineering)

Assignment No.: 01

Due Date: Friday, 15 May 2026




Submitted in partial fulfilment of the requirements for MAT1581
at the University of South Africa.

,UNISA | MAT1581 Assignment 01 — Due: 15 May 2026



Question 1: Binomial Theorem — Finite Expansions


Question 1.1


Question: Expand using the binomial theorem:


(2x − 3y)5



Solution:

The binomial theorem for (a − b)5 gives:


(a − b)5 = a5 − 5a4 b + 10a3 b2 − 10a2 b3 + 5ab4 − b5



Let a = 2x and b = 3y. Substitute into the formula:


(2x − 3y)5 = (2x)5 − 5(2x)4 (3y) + 10(2x)3 (3y)2 − 10(2x)2 (3y)3 + 5(2x)(3y)4 − (3y)5



Calculate each term one at a time.

Term 1:
(2x)5 = 25 x5 = 32x5


Term 2:
−5(2x)4 (3y) = −5 · 24 x4 · 3y = −5 · 16x4 · 3y = −240x4 y


Term 3:
10(2x)3 (3y)2 = 10 · 8x3 · 9y 2 = 720x3 y 2


Term 4:
−10(2x)2 (3y)3 = −10 · 4x2 · 27y 3 = −1080x2 y 3


Term 5:
5(2x)(3y)4 = 5 · 2x · 81y 4 = 810xy 4




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, UNISA | MAT1581 Assignment 01 — Due: 15 May 2026


Term 6:
−(3y)5 = −35 y 5 = −243y 5


Therefore:


(2x − 3y)5 = 32x5 − 240x4 y + 720x3 y 2 − 1080x2 y 3 + 810xy 4 − 243y 5



Question 1.2


Question: Expand using the binomial theorem:


4
x3 + y 2



Solution:

The binomial theorem for (a + b)4 gives:


(a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4



Let a = x3 and b = y 2 . Substitute:


(x3 + y 2 )4 = (x3 )4 + 4(x3 )3 (y 2 ) + 6(x3 )2 (y 2 )2 + 4(x3 )(y 2 )3 + (y 2 )4



Calculate each term.

Term 1:
(x3 )4 = x3×4 = x12


Term 2:
4(x3 )3 (y 2 ) = 4x9 y 2


Term 3:
6(x3 )2 (y 2 )2 = 6x6 y 4


Term 4:
4(x3 )(y 2 )3 = 4x3 y 6


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