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Answer:
GIVEN: Side of square (4q - 5r + 3s) & portion of the paper with an area of (2q² – 12qr + 3 qs + 8r² – 5rs + s²) to be removed
UNKNOWN: total area of the paper & remaining area of the paper
FORMULA:
for the area of the paper : multiplication
Area of paper = (4q - 5r + 3s)×(4q - 5r + 3s)
for the remaining area : substraction
16q² + 25r² + 9s² - 40qr + 24qs - 30rs
- 2q² + 8r² + s² - 12qr + 3qs - 5rs
CONCLUSION: the remaining area of the paper is (14q² + 17r² + 8s² - 28qr + 21 qs - 25rs)
SOLUTION:
Area of paper = (4q - 5r + 3s)×(4q - 5r + 3s)
= 16q² + 25r² + 9s² - 40qr + 24qs - 30rs
16q² + 25r² + 9s² - 40qr + 24qs - 30rs
- 2q² + 8r² + s² - 12qr + 3qs - 5rs
14q² + 17r² + 8s² - 28qr + 21 qs - 25rs