Class 7 I. Special Factoring A. Review of Basic Factoring B. Factoring Sums and Differences of Squares C. Factoring Sums and Differences of Cubes II. Solving Quadratic Equations A. Factoring B. Word Problems C. Practice Problems I. Special Factoring A. Review of Factoring ax^2 + bx + c This week I want to pick up where we left off last week, with factoring quadratics. I want to start by reviewing how we factor expressions of the form ax^2 + bx + c. Let's consider an example of something we would like to factor: (on board) x^2 - 3x - 10 Someone who has a good memory: tell me how we go about factoring this expression. (Let class walk through it.) We make our standard list of possibilities for -10: (on board) Possibilities for -10: 1,-10 -> (x+1)(x-10) -> x^2 - 9x - 10 -1,10 -> (x-1)(x+10) -> x^2 + 9x - 10 2,-5 -> (x+2)(x-5) -> x^2 - 3x - 10 -2,5 -> (x-2)(x+5) -> x^2 + 3x - 10 So the answer is (x+2)(x-5) Notice that the number in the middle, the coefficient of x, always comes out to be the sum of the two numbers we're trying out. This can be a bit of a shortcut for factoring things that look like x^2 + bx + c: rather than multiplying everything out, we just have to check the sums of all the pairs. Unfortunately, this trick doesn't work so neatly when the coefficient of the x^2 is not a 1. B. Factoring Sums and Differences of Squares Before we move on, I'd like to point out one special case of factoring ax^2 + bx + c: the case when b, the coefficient of the x term, is zero. Consider (on board) x^2 - 9 How can we factor this, using our standard procedure? (let class answer) (on board) Possibilities for -9: 1,-9 -1,9 -3,3 Using our trick, we note that we're looking for a pair of numbers that add up to the coefficient of x -- which is zero in this case. Looking at this list, we can see that the answer has to be -3, 3, since only they add up to 0. Thus, the answer is (on board) (x-3)(x+3) We can see a pattern that's going to be true whenever we have a difference of squares, something like (on board) x^2 - a^2 The answer is just going to be: (on board) (x - a)(x + a) Thus, for example, what is the factorization of (on board) x^2 - 64 (let class answer) (on board) (x - 8)(x + 8) Suppose we consider a closely related problem: (on board) Factor x^2 + 9 If we use our standard method, we list the possibilities for 9: (on board) Possibilties for 9: 1,9 -1,-9 3,3 -3,-3 We're looking for a pair of numbers that add up to zero, but we can see immediately that can never happen! Since we have to get a positive 9 when we multiply our possibilities, the possibilities always have to have the same sign -- both positive or both negative. There is therefore no way to add the numbers and get zero, so this expression cannot be factored. Thus, we see that a sum of squares, something of the form (on board) x^2 + a^2 cannot be factored. C. Factoring Sums and Differences of Cubes We can also find tricks for factoring sums and differences of cubes. A little quick arithmetic will verify that: a^3 + b^3 = (a + b) (a^2 - ab + b^2) a^3 - b^3 = (a - b) (a^2 + ab + b^2) Let's check these two rules: (a+b) (a^2 - ab + b^2) = a^3 - a^2 b + ab^2 + a^2 b - ab^2 + b^3 = a^3 + b^3 (a-b) (a^2 + ab + b^2) = a^3 + a^2 b + ab^2 - a^2 b - ab^2 - b^3 = a^3 - b^3 Generally the quadratic that you get by this factorization cannot itself be factored, so the first step is the only one you can do. Example: x^3 + 27 Here, x^3 is clearly the cube of x, and 27 is 3^3. So factoring: x^3 + 27 = (x + 3)(x^3 - 3x + 9) If you try some numbers, you'll see quickly that x^2 - 3x + 9 is not factorable. We can verify that using the discriminant: a = 1, b = -3, c = 9, so b^2 - 4ac = (-3)^2 - 4(1)(9) = 9 - 36 = -27 This is a negative number, so the quadratic cannot be factored. II. Quadratic Equations A. Solution by Factoring Why have we spent all this time learning to factor? The main reason is because factoring lets us solve one of the most common types of equations: quadratic equation. A quadratic equation is just something of the form ax^2 + bx + c = 0, or something that can be put into that form. For example, 3x^2 = 7x - 2 is a quadratic equation, because we could re-write it as 3x^2 - 7x + 2 = 0. To solve this equation, the first thing we do is factor the quadratic. How do we do that? (let class walk through it) (on board) Possibilities for 2: 1,2 -1,-2 Possibilities for 3: 1,3 (on board) (3x - 1)(x - 2) = 0 So we've replaced the expression involving x^2 by a product of two other expressions. So what good is that? To proceed from here, we make a clever observation: the only way I can multiply two numbers and get zero is if one of those numbers is zero. In this expression, we have (3x - 1) multiplied by (x - 2). We know that when we multiply them, we get zero. This is only possible if either (3x - 1) is zero or (x - 2) is zero. Thus, we can write: (on board) 3x - 1 = 0 or x - 2 = 0 These a simple linear equations, and we can solve them immediately. (on board) 3x - 1 = 0 or x - 2 = 0 +1 +1 +2 +2 3x = 1 or x = 2 x = 1/3 or x = 2 Thus we've solved for x, and found two possible solutions. We can check our answers by plugging into the original equation: (on board) 3x^2 - 7x + 2 = 0 3(1/3)^2 - 7(1/3) + 2 = 0 1/3 - 7/3 + 2 = 0 0 = 0 (on board) 3x^2 - 7x + 2 = 0 3(2)^2 - 7(2) + 2 = 0 12 - 14 + 2 = 0 0 = 0 So both our solutions check out. We can see that, in general, there will be two solutions to quadratics. This gives us a general procedure for solving equations involving quadratics: (on board) 1. Rearrange the equation so that one side is zero and the other side is in the form ax^2 + bx + c. 2. Factor ax^2 + bx + c. 3. Set each of the two factors equal to zero. 4. Solve the resuling linear equations to get the solutions. B. Word Problems Of course there are word problems for this type of equation. One common example is problems involving areas. Here's an example: A farmer is putting up a fence around a field. The field has an area of 15,000 m^2, and is 50 m longer than it is wide. How many meters of fencing does he need to buy? For a problem like this, step 1 is always drawing a picture: 1. Draw a Picture ______________________ | | | | | | w | | |____________________| w + 50 Next we compute the area in terms of the variables. 2. Compute the Area Area = w (w + 50) The next step is write an equation involving the area: 3. Write an equation w (w+50) = 15,000 4. Solve the equation w^2 + 50w = 15,000 w^2 + 50w - 15,000 = 0 (w + 150)(w - 100) = 0 w = -150, 100 5. Choose the meaninful solution In this case, a width can't be negative, so w = 100 m is the meaningful solution. 6. Answer in English We were asked to find the amount of fencing, which means the perimeter. The width we've already found: it's 100 m. The length is w+50 = 150 m. So the perimeter is 2*(length + width) = 2*(100+150) = 500. So the answer is: "The farmer needs 500 m of fencing." Another example is problems involving motion with acceleration. If an object moves at a constant velocity v, the distance it covers is just v times the time -- distance = rate times time. It turns out there is a similar formula for an object travelling at a constant acceleration. The distance travelled by an object moving at constant acceleration a and starting at rest is 1/2 times a times the time squared: 1/2 a t^2. The origin of the formula is a topic for a physics class, not an algebra one, but because of that formula quadratics come up frequently when there is motion with constant acceleration. An example of motion of this sort is falling. Here's an example problem using this: (on board) A ball is thrown upward at 10 meters per second from the edge of a 75 meter cliff. Its height after t seconds is 75 + 10t - 5t^2. Find the time when (a) the ball falls past its original height and (b) the ball hits the ground below the cliff. We approach this in a way very similar to a geometry problem. First, we draw a picture. (on board) 1. Draw a picture. (draw picture on board) Next, we use the information given to write down an equation. (on board) 2. Write an equation using the information given. In this case, we know the height at every time -- we have a formula for it. First, we want to know when the ball passes its original height. What equation can we write to describe this situation? (let class answer) In this case, the ball is at 50 meters, so we write an equation: (on board) 75 + 10t - 5t^2 = 75 Now we just need to solve. (on board) 3. Solve the equation. How do we solve this? (let class answer) (on board) 75 + 10t - 5t^2 = 75 0t - 5t^2 = = 0 5t (2 - t) = 0 Note that this is an unusual sort of quadratic: there is no constant term. In this case, we can factor without needing to try lots of pairs of numbers. However, we still arrive at something times something else, so we still use the principle that if we multiply two numbers and get zero, one of the things we multiplied must be zero already. Thus: (on board) 5t = 0 or 2 - t = 0 t = 0 or t = 2 Finally, we have (on board) 4. Choose the solution that makes sense and write the answer in English. In this case, which solution makes sense? (let class answer) The solution t = 0 is just telling us that, after zero seconds, before the ball has gone anywhere, it's at a height of 50 meters. That's not particularly interesting, since we knew that already. We want to know after how many seconds the ball passes 50 meters on the way down. Thus, we take the answer 2. (on board) The ball passes its original height after 2 seconds. Now let's try part (b) of this problem. How would we do this? (let class walk through it) The equation is (on board) 75 + 10t - 5t^2 = 0 We can simplify by dividing by -5: (on board) t^2 - 2t - 15 = 0 Again, we can factor: (t - 5)(t + 3) = 0 t = 5, -3 What answer makes sense here, physically? (let class answer) Clearly we want the positive solution, since we know the ball has to hit the ground after we throw it, not before. Thus, the solution is (on board) The ball hits the ground after 5 seconds C. Practice Problems Pratice, as time permits: Solve: 1 = 3x - x^2 Two years ago, one boy was twice as old as his brother. The product of their ages was 72. How old is each one now?