Rearranging the conditional constraint gives x – 2y – 6 = 0. The polynomial can be formatted as x cube + (-2y) cube + (-6) cube – 3(x)(-2y)(-6), which evaluates directly to 0 because the base sum is 0.
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Factoring n cube – n gives the expression (n – 1)(n)(n + 1), which represents three consecutive integers. In any three consecutive numbers, at least one is a multiple of 2 and one is a multiple of 3.
We divide the given quadratic area expression by the given width binomial. Factoring the numerator 2x square + 7x + 3 yields (2x + 1)(x + 3) and cancelling the common width factor leaves the length.
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).
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.