Write an article based on this "Count the number of negative signs in your multiplication problem. Decide the sign of your answer based on the number of negative signs in the problem. Multiply numbers from 1 - 10 using basic times table knowledge. If necessary, break larger numbers into manageable chunks. For more difficult numbers, use long multiplication."
A basic multiplication problem between two or more positive numbers will always result in a positive answer. However, each negative sign added to a multiplication problem flips the sign from positive to negative or vice versa. To begin an integer multiplication problem, count the number of negative signs in the problem. Let's use the example problem -10 5 -11 -20. In this problem, we can clearly see three negative signs. We'll use this information in the next step. As noted above, the answer to a multiplication problem involving only positive integers will be positive. For each negative negative sign in your problem, flip the sign of your answer. In other words, if your problem has one negative sign, your answer will be negative; if it has two, your answer will be positive, and so on. A good rule of thumb is that odd numbers of negative signs give negative answers and even numbers of negative signs give positive answers. In our example, we have three negative signs. Three is an odd number, so we know our answer is negative. We can put a negative sign in the space for our answer, like this: -10 5 -11 -20 = -__ The product of any two numbers lesser than or equal to 10 is covered in basic times tables (see above). For these simple cases, just write the answer. Remember that, in problems that only use multiplication signs, you can move the integers around so that you're able to multiply simple numbers with each other. In our example, 10 5 is covered in the basic times table. We don't have to account for the negative sign on the ten because we've already found the sign of our answer. 10 5 = 50. We can insert this into our problem like this: (50) -11 -20 = -__ If you're having difficulty visualizing basic multiplication problems, think of them in terms of addition problems. For instance, 5 10 is like saying "five, ten times." In other words, 5 10 = 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5. If your multiplication problem involves numbers greater than ten, you don't necessarily have to use long multiplication. First, see if you can break one or more of your numbers down into smaller, more workable pieces. Since, with basic times table knowledge, you can solve simple multiplication problems almost instantly, breaking a difficult problem into several of these easy problems is usually simpler than solving the single difficult problem. Let's look at the second half of our example problem, -11 -20. We can omit the signs because we've already figured out the sign of our answer. 11 20 looks intimidating, but if we rewrite the problem as 10 20 + 1 20, suddenly, it's much more manageable. 10 20 is just 2 times 10 10, or 200. 1 20 is just 20. Adding up our answers, we get 200 + 20 = 220. We can re-insert this into our problem as follows: (50) (220) = -__ If your multiplication problem involves two or more numbers greater than 10 and you're not able to find the answer by dividing your problem into workable chunks, you can still solve via long multiplication. In long multiplication, you line your answers up as you would in an addition problem and multiply each digit in the bottom number by each digit in the top number. If the bottom number has more than one digit, you'll need to account for digits in the tens, hundreds, and so on place by adding zeroes to the right side of your partial answer. Finally, to get your final answer, add up all the partial answers. Let's return to our example problem. Now, we must multiply 50 by 220. This will be difficult to break into easier chunks, so let's use long multiplication. Long multiplication problems are easier to keep track of if the smaller number is on the bottom, so let's write our problem with 220 on top and 50 on the bottom. First multiply the digit in the ones place of the bottom number by each digit of the top number. Since 50 is on the bottom, 0 is the digit in the ones place. 0 0 is 0, 0 2 is 0, and 0 2 is zero. In other words, 0 220 is zero. Write this below your long multiplication problem in the ones place. This is our first partial answer. Next, we'll multiply the digit in the tens place of our bottom number by each digit of the top number. 5 is the digit in the tens place of 50. Since this 5 is in the tens place, rather than the ones place, we write a zero below our first partial answer in the ones place before proceeding. Next, we multiply. 5 0 is 0. 5 2 is 10, so write 0 and add one to the product of 5 and the next digit. 5 2 is 10. Normally, we would write 0 and carry the 1, but in this case we also add the 1 from the previous problem, giving us 11. Write down "1". Carrying the 1 from the tens place of 11, we see that we're out of digits, so we just write it to the left of our partial answer so far. Recording all this, we're left with 11,000. Next, we just add. 0 + 11,000 is 11,000. Since we know the answer to our original problem is negative, we can safely say that -10 5 -11 -20 = -11,000.