Let a, b, c, d Є C, where C is the field of complex numbers.
Let m, n, p, q Є N, where N is the field of natural numbers, including 0.
If w, x, y, z Є C are unknown, the product of the two binomials (awm + bxn) and (cyp + dzq) is equal to the following:
acwmyp + adwmzq + bcxnyp + bdxnzq.
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Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
FOIL. First terms Outer terms Inner terms Last terms
No. A counter-example proves the falsity: Consider the two binomials (x + 2) and (x - 2). Then (x + 2)(x - 2) = x2 - 2x + 2x - 4 = x2 - 4 another binomial.
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
bi- = 2 Binomials have two terms.