Factoring a Degree 2 PolynomialDate: 3/21/96 at 0:40:13 From: Brandon Subject: Math I don't get this. I have tried, but not succeeded! I am in grade nine. Questions like x exponent2 - 12x - 85 I don't get these factor things ! Please help me a.s.a.p Thanks, Brandon Date: 3/21/96 at 13:45:12 From: Doctor Ethan Subject: Re: Math Hey Brandon, Well, factoring can be tricky. There are a few tricks. Let's see if we can use them. I will use x^2 to mean x exponent2, okay? So we have x^2 - 12x - 85 Here is the idea. We want to write this as the product of two binomials that look like (x + a)(x + b) where a and b are any number. Right? So we want to figure out what a and b have to be. Well, let's work backwards for a second. (x + a)(x + b) = x^2 + ax + bx + a*b = x^2 + (a+b)x + a*b Right? Well then, if we have the equation that you gave us, we know two things. 1. (a+b) = -12 2. a*b = -85 Let's work with number 2 first. If a times b is -85, what can a and b equal? Let's make a chart. a | b ----------- 1 | -85 5 | -17 17 | -5 85 | -1 Now, which one of these possibilities gives us a + b = -12? Well, let's try them. 1 + (-85) = -84 Nope. 5 + (-17) = -12 Yep. 17 + (-5) = 12 Nope. 82 + (-1) = 84 Nope. Only one of them works, so (x + 5)(x + (-17)) = (x + 5)(x - 17) is the answer. Hope that this helps. -Doctor Ethan, The Math Forum Date: 3/21/96 at 14:31:12 From: Doctor Terrill Subject: Re: Math Dear Brandon, Factoring isn't an easy thing to figure out. Usually, the answer isn't obvious, and you have to think a lot about the problem before you can figure it out. I'll try to explain how I think about factoring. First of all, instead of x exponent 2 I'll write x^2: the caret symbol means that the 2 should be up above the x, like an exponent. When you factor a number, like 12, you write it as a product of prime numbers: for instance, 12 = 2 * 2 * 3 (* means times). To factor x^2 - 12x - 85, we want to write it as a product of other algebraic expressions. In other words, when we factor x^2 - 12x - 85, we will be able to write it as the product of two or more algebraic expressions. So, we want to figure out what a, b, c, and d will have to be to make the equation (a + b) (c + d) = x^2 - 12x - 85 true. First, let's multiply out (a+b) (c + d). A good way to remember how to do this is to think of (a + b) as one number at first. For instance, 7 * (c + d) = 7 * c + 7 * d. So, if we think of (a + b) as one number, we get (a + b) (c + d) = (a + b) * c + (a + b) * d. Now, we can multiply out (a + b) * c and (a + b) * d. This gives us ac + bc + ad + bd. So, we want to figure out what a, b, c, and d have to be to make the equation x^2 - 12x - 85 = ac + bc + ad + bd. Since we have an x^2 on the left-hand side, we know that something on the right will probably equal x^2. Let's suppose that ac = x^2. We know that ac = x^2 when a = x and c = x. We'll try putting these values into our equation, and see what happens. (We could also have a = -x and c = -x, or a = x^2 and c = 1, but it's most likely that a = x and c = x.) Substituting a = x and c = x into our equation gives us x^2 - 12x - 85 = x ^2 + bx + ax + bd. We can factor the expression on the right by taking the x out of bx + ax. This gives us x^2 - 12x - 85 = x^2 + (b + d)x + bd. Since we have a -12x on the left, we know that (b + d)x will have to equal -12x. Also, since -85 is the only part of the equation on the left that doesn't have an x in it, bd will have to equal -85. ( x^2 has an x in it, as does (b+d)x, so those can't equal -85.) So, we know that b + d = -12, and bd = -85. We need to find an a and b that work in both these equations. One way to do this is by making a list of all the b's and d's that work in the equation bd = -85: -85 = -1 * 85 = -85 * 1 = -17 * 5 = -5 * 17. If all we wanted was b and d such that bd = -85, then any of these numbers would work: for instance, we could have b = -1 and d = 85. But we have other restrictions on b and d: they have to add up to -12. That only works for one set of numbers, -17 and 5, because -17 * 5 = -85, AND -17 + 5 = -12. So, b is -17 and d is 5. We can substitute these into x^2 - 12x - 85 = (a + b) (c + d). This gives us: x^2 - 12x - 85 = (x - 17) (x + 5). We can check if this is the correct answer by multiplying it out: (x - 17) (x + 5) = (x - 17) * x + (x - 17) * 5 = x^2 - 17x + 5x - 17 * 5 = x^2 - 12x - 85, which is what we wanted. So the factorization of x^2 - 12x - 85 is (x - 17) (x + 5). It took a lot of steps to find the answer, but it gets easier the more you do it. Here's one for you to try: Try factoring x^2 + x - 6. Remember, we want to be able to write it in the form (a + b) (c + d), and since we have an x^2 in the equation, a and c are probably both x. So you want to find b and d in the equation (x + b) (x + d) = x^2 - 12x - 85. Try taking it from there. I hope this helped, -Doctor Terrill, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/