Method to solve binomials of second degree, Mathematics

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In this part we look at another method to obtain the factors of an expression. In the above you have seen that x2 - 4x + 4 = (x - 2)2 or (x - 2)(x - 2). If you observe it carefully we find that the middle number - 4 is the sum of -2 and -2 and the last term 4 is the product of -2 and -2. That is, if you think that so and so number might be the factors of the binomial those numbers should satisfy this condition. We take another example. You are given x2 + 15x + 56 and asked to factorize it. Now if you think that, say, 6 and 7 are the factors of this expression then their product should be equal to 56 and their sum should be equal to 15. However in this case we observe that the product is 42 and the sum is 13. Therefore, 6 and 7 cannot be the factors of this expression. Now try 7 and 8. We find that their product is 56 and the sum 15. That is, 7 and 8 are the factors of the given expression. This can be clarified by multiplying (x + 8) and (x + 7).

One point to which we have to pay attention is that we have to take even signs into consideration. For instance, consider an expression x2 - 17x + 70. What could be the factors of this expression? Let us try 7 and 10. No doubt, the product is 70 and the sum 17. Still these cannot be the factors of the given expression, because the sum is -17 and we got only 17. Now let us try -7 and -10. The sum of these two numbers gives us -17 and their product as 70. This is what we require. Therefore, the factors are x - 7 and x - 10 (observe that in this case if we took x = -7 and         x = -10, we would have got the factors as x + 7 and x + 10, whose multiplication would give us x2 + 17x + 70 and not x2 - 17x + 70. That is, the values should be considered as they are). Now let us consider an expression x2 - 3x - 70. Let us try 7 and -10 for this expression. The sum of these two values is -10 + 7  = -3 and the product being -70. That is, x + 7 and x - 10 are two factors of the given expression and not x - 7 and x + 10.


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