The total outer side length including the path becomes 40 + 2s metres. Subtracting the inner square area (1600) from the expanded outer square area provides the final algebraic expression for the path.
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We factor out the common algebraic term 3p from the expression. The remaining quadratic part inside the brackets is s square – 5s + 4, which is split into the linear binomial factors (s – 1) and (s – 4).
We first take out the common constant factor 6 from the volume expression. This leaves a square – 4b square inside brackets, which breaks down further into (a + 2b)(a – 2b) using the difference of squares formula.
We apply the difference of squares identity, x square – y square = (x + y)(x – y). Rewriting 36s square as (6s) square and 49t square as (7t) square directly provides the two unique linear dimensions.
The area expression fits the identity template of x square – 2xy + y square. Since 25a square is (5a) square and 9b square is (3b) square, it factors into the product of two identical binomial dimensions.