In Part IV of this series, we examine how we solve factorable quadratics of the form ax^2 - bx - c, in which both the b and c terms are negative. Such an example would be x^2 - 4x - 5. This subclass of quadratics, which we will call the “bc-negatives, ” are easily solvable using the factoring techniques we learned about in Parts II and III of this series of articles.
Once again, let’s get right to the point and solve the quadratic in the first paragraph: x^2 -4x -5. Remember, we are looking for the solutions, or those values of x which when plugged in for this variable make the quadratic zero. This is another way of saying that those values are the zeros of the function, or those values are the ones which make the quadratic vanish. As in Part III, we look to see the factors of the c-term, which in this example is -5. When considering the factor pairs, we will disregard the negative sign and only examine the factors of 5. Since 5 is a prime number, its only factors are 1 and itself.
Remember: the c-term of the quadratic is always formed by multiplying the numbers of the factor pair. Since the c-term is negative, the 1 and 5 must be of opposite sign, since this is the only way we can get a negative product. Thus we must have 1 with a -5 or -1 with a 5. Both (1)(-5) and (-1)(5) = -5. Since the b-term is obtained by adding and since the b-term is negative, the larger number of the factor pair must be negative. If this were not true, then the result would be positive. You can think of this as the larger negative force overpowering the smaller positive force to give a net negative force. You can also relate this back to chemistry in that if you have more negative electrons than positive protons you end up with a net negative charge, and vice versa. (You see how you can relate math back to many other things in the real world. ) Consequently, this quadratic must factor as (x - 5)(x + 1) and the solutions are 5 and -1. Once again, you can plug either of these two values back into the original quadratic to show that they make the equation equal to zero.
To see how this method works with another example, let’s try x^2 - 4x - 12. The factor pairs of 12 are 1-12, 2-6, and 3-4. The only pair that will combine to give 4 is 2-6. Now we have to make an adjustment for the b-term. We know that (2)(-6) and (-2)(6) = -12. Of these two examples, the greater negative force occurs in the pair (2)(-6) to yield a net negative of -4. This is what we need, and hence x^2 - 4x - 12 = (x + 2)(x - 6), and the answers are -2 and 6. Each of these will make the quadratic vanish or become equal to zero upon substitution.
In the next article, we will look at the final case of factorable quadratics, ones of the form x^2 + bx - c. With all the tools we now have available, we will be able to dispatch with this last class with the blink of an eye. In fact, if you’ve followed this series this far, you already have the knowledge to do so now. Stay tuned. . .
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