By factor theorem, substituting x = 2 and x = 1/2 both make the polynomial equal 0. Equating the resulting expressions 4p + 10 + r = 0 and p/4 + 5/2 + r = 0 simplifies perfectly to prove ...
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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.
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 polynomial matches the identity template a square – 2ab + b square. Since 16s square is (4s) square and 25t square is (5t) square, it condenses directly into (4s – 5t) square.
We express the product as (20 – 2) (30 – 1) and expand using the distributive property. Evaluating the separate numerical products step by step gives the clean final total.