How do you solve p² equals 2p plus 99 by factoring?

Answer 1

First you rewrite the equation so everything is on one side. You do this by subtracting (2p + 99) from both sides. The new equation is: p2 - 2p - 99 = 0. This is called a "trinomial", or more specifically, a "quadratic equation". Now you can factor it into two "binomials". First write: (p + __)(p + __) where the blanks represent constants. Then you try to fill in the blanks with something that works. Once you have something you think might work, you multiply the two binomials together (by the FOIL method - see below if you don't know the FOIL method) and see what you get. If you get p2 - 2p - 99, then you were right. When trying to figure out what goes in the blanks, remember that the two terms in the blanks, multiplied together (the L part of FOIL), will be equal to the last term in your trinomial, in this case, -99. So what sets of two numbers can we multiply together to get -99? First of all, we know, in general, that in any set of two numbers that, when multiplied together, result in a negative number, one of them must be positive and the other one negative. For example, -1 and 99. 1 and -99 also works, but 1 and 99 will not work, nor will -1 and -99. So, you need to construct a set of all possible pairs of factors that multiply to -99 (you don't really have to do this in all cases - you COULD just try each pair as you think of it, and only look for another pair after the first one fails). The following is a complete list of all pairs of factors that multiply to -99: -1, 99

1, -99

-3, 33

3, -33

-9, 11

9, -11 But which one of these is the answer? You just have to try them out. But, you can often narrow it down a little with some common sense. You know that the middle term of your trinomial (in this case -2p) is the sum of cross-products (the variable term in one binomial times the constant term in the other binomial, and vice versa, or the O and I parts of FOIL). Now, this would get complicated if the coefficient of the x2 term was something other than 1, but since it isn't, this is rather straightforward. Basically, you're looking for two numbers that, when added together, result in -2. 9 and -11 look like a good bet, so we'll try them first, by filling them in the blanks in the incomplete binomials: (p + 9)(p - 11) Then, you just multiply the two binomials using the FOIL method: F (Product of the First terms in each binomial): p x p = p2

O (Product of the Outside terms in each binomial): p x -11 = -11p

I (Product of the Inside terms in each binomial): 9 x p = 9p

L (Product of the Last terms in each binomial): 9 x -11 = -99 F + O + I + L = p2 + -11p + 9p + -99 = p2 + (-11p + 9p) - 99 = p2 - 2p - 99 which is our original quadratic equation. And so, we have successfully factored the quadratic equation as (p + 9)(p - 11). But we haven't solved it yet. To solve it, we need to go back to the trinomial. Remember that we started with: p2 - 2p - 99 = 0 Now, we know, from our factoring of this equation, that p2 - 2p - 99 = (p + 9)(p - 11). So we can now say that: (p + 9)(p - 11) = 0 Now what do we know about any two numbers that, when multiplied to each other, result in zero? We know that at least one of the two numbers must be zero. Therefore, we know that it must be true that either (p + 9 = 0) or (p - 11 = 0). If either of these statements is true, then it is true that (p + 9)(p - 11) = 0, and therefore that p2 - 2p - 99 = 0. Therefore, p + 9 = 0 or p - 11 = 0 defines the complete set of solutions to our equation. And all we have to do is solve the two simple equations for the variable p. When is it true that p + 9 = 0? There is only one value that we can substitute for p that makes this true. And that value is -9. Algebraically, this is shown by subtracting 9 from both sides of the equation. On the left side, p + 9 becomes p + 9 - 9, which simplifies to simply p. On the right side, 0 becomes 0 - 9, which simplifies to -9, and we are left with p = -9. And so, p = -9 is one of the two solutions. Similarly, the only time it is true that p - 11 = 0 is when p = 11. That is the second solution. The two solutions to this equation are therefore: p = -9 and p = 11 It is often helpful to think of an equation graphically. Any equation whose highest power is a 2 (a squared variable) will, graphically, take the form of a parabola. A parabola looks sorta like the letter U, but the sides are not quite vertical. The sides do become increasingly steep as you move away from the center, but they never quite reach totally vertical. The point behind this explanation is that, if you try to draw a parabola on a graph, there are a limited number of ways you can draw it. You can draw it rightside up, or you can draw it upside down. You can draw it with the vertex (the bottom-most point of the U, or the topmost point if you draw it upside down) either above the x axis, or below the x axis, or even exactly on the x axis. You can draw it very narrow or very wide, or any width in between. But here's the thing. If your vertex is below the x axis, and the parabola is rightside-up, no matter how wide or narrow you draw your parabola, it will always cross the x axis in exactly two places. The same is true if you draw an upside-down parabola with the vertex above the x axis. That is why a quadratic equation, if it has any solutions at all, has two solutions. If you draw a rightside-up parabola with the vertex above the x axis (for example, replace the -99 in the original trinomial with a positive 99), or an upside-down parabola with the vertex below the x axis, it will never cross the x axis at all, no matter how far you extend the legs, because it's always getting farther away from the x axis, not closer. If you had such an equation, one that, if graphed, would never cross the x axis, then you would find it impossible to factor this equation. You could try every pair of values for the constants, even non-integer values, even non-rational values, and you would never get a solution. You could even plug the coefficients into the quadratic formula [-b ±√(b2 - 4ac)]/2a, and you would find that the term (b2 - 4ac) under the square root operator was negative, and as we all know, negative numbers do not have square roots (or at least not realsquare roots), and so there is no real solution to the quadratic formula. And this makes sense, because the x axis is defined as the line y = 0. If a graph never crosses the x axis, then it is never true that y = 0. But if y is never equal to zero, then the equation is never equal to zero, so there are no values of x that can result in the equation being zero. The only other possibility is that the vertex of the parabola is exactly on the x axis. For example, take the formula y = x2 - 4x + 4. If you graph this formula, you would find that it touches the x axis at only one point, the very lowest point of the parabola, where x = 2, which is the vertex of the parabola. This parabola would appear to violate the rule I stated in the previous paragraph, that if a quadratic equation has any solutions at all, it always has exactly two solutions. Because this quadratic equation, apparently, has only one solution, x = 2. But, if you were to factor this equation, you would get (x - 2)(x - 2). The two factors, in this case, are identical, and therefore are both solved by the solution x = 2. There are still two solutions to this quadratic equation. They just happen to be identical. So the rule is intact. To slightly restate that rule: Every quadratic equation (any equation whose highest order is 2) will have either zero or two real solutions, never 1, and never more than 2.