Exploring Algebraic Identities
Using factorisation
Use factorisation to solve problems about rectangles, boxes and pools.
Factorising turns an area into length × breadth. Saira has one x² piece, eight x-strips and fifteen unit squares: area x² + 8x + 15 = (x + 3)(x + 5). So her rectangle is x + 5 by x + 3.
Example 18. A pool's breadth is 4 m less than its length, and its area is 96 m². With length x: x(x – 4) = 96, so x² – 4x – 96 = 0. Split –4x as –12x + 8x: (x – 12)(x + 8) = 0.
If a product is zero, one of the factors is zero. So x = 12 or x = –8. A length cannot be negative, so the pool is 12 m by 8 m. Always check which answers make sense in the story.
Volumes factor into three parts: 3ps² – 15ps + 12p = 3p(s – 1)(s – 4).
A factor x – k means the expression is 0 at x = k. That is how the book's Problem 10 finds p and r.
A neat proof. n³ – n = (n – 1)n(n + 1): three numbers in a row. One of them is even and one is a multiple of 3, so the product is always a multiple of 6.
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