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This question shows up often on standardized tests (SAT, ACT, GRE, etc) and can be solved in about two seconds 99.9% of the time if you memorize your square tables up to about 25 or so. Here's how to solve it: 1. Add 1 to the given product. 2. The answer is the square root of this number -/+ 1. Why it works: Any pair of consecutive odd or even numbers can be represented by X and X + 2, but also X - 1 and X + 1. We use this latter representation because it is in conjugate pair form, and the product will be X^2 - 1. So, all you are doing is reversing the process. Step 1 eliminates the -1 term. Taking the square root leaves you with X, a number which will be 1 greater and 1 less than the two numbers you are asked for in the original problem. Knowing this, you can memorize a list of common consecutive odd/even integer products, which will also equal the number in between^2 - 1: 11 x 13 = 12^2 - 1 = 143 12 x 14 = 13^2 - 1 = 168 13 x 15... read more

When multiplying polynomials of higher degree than quadradics (cubics * cubics or cubics * quartics, for example) instead of writing out each term using the distributive property, you can make a simple 2 x 2 matrix with the coefficients of one equation in a row and the other in a column. You are still using the distributive property multiplying each term by every other term, but visually it is easier to keep track of like terms because you can sum them up in diagonals rather than picking them out amongst a long string of numbers and letters. For example: (2x^3 + x^2 - 4x + 1) * (x^4 + 5x^3 + 2x - 5) = 2 1 -4 1 Diagonals start at 2 go to down to -10, then over to -5 ------------------ 1 | 2 1 -4 1 5 | 10 5 -20 5 0 | 0 0 0 0 2 | 4 2 -8 2 -5|-10 -5 20 -5 Adding diagonals (summing all terms lower left to upper right) gives you: 2x^7 + 11x^6 + x^5 - 15x^4 - 3x^3 - 13x^2 + 22x - 5 Note that the x^2 term is missing... read more

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