Consider the following problem:<\/p>\n
Reduce 55\/528 to lowest terms.<\/p>\n
Of course, the key is knowing the number(s) that divide into both the top and the bottom. In this case, 11 does:<\/p>\n
55\/528=(5×11)\/(48×11) = 5\/48<\/p>\n
Since knowing that 11 divides into a number is obviously useful, how do we do so?<\/p>\n
Multiply the ones digit (8, in the case of 528) by 1, then the tens digit (2, in 528) by -1, then the hundreds digit (5) by 1. Add up the three products: 8(1) +2(-1) + 5(1) = 11. If the sum is divisible by 11, the number is. Since, for 528, the sum is 11, 528 is divisible by 11.<\/p>\n
231 is also divisible by 11: 1(1) +3(-1) + 2(1) = 0. Since 0 is divisible by 11 (0\u00f711=0), 231 also is. Indeed, 231=11(21).<\/p>\n
The pattern continues for numbers in the thousands. Consider 1936. 6(1) + 3(-1) + 9(1) + 1(-1) = 11. Therefore, 1936 is divisible by 11. In particular, 1936=11(176).<\/p>\n
Now consider a polynomial factoring example:<\/p>\n
Factor x2<\/sup> – 4x – 165<\/strong><\/p>\n Solution: From my post on easy trinomials<\/a>, you know that we must find two numbers that multiply to make -165, but add to make -4. Knowing by sight that 11 divides into 165, we easily recognize our numbers as 11 and -15.<\/p>\n x2<\/sup> – 4x – 165 = (x+11)(x-15).<\/p>\n I’ll have more to say about divisibility in future posts:)<\/p>\n