Mastering Math: Rectangles, Primes, And Cubes
Hey math whizzes and curious minds! Today, we're diving deep into some awesome math problems that will get your brain buzzing. We've got a cool mix of geometry, number theory, and algebra lined up for you. So grab your pencils, get comfy, and let's break down these challenges step-by-step. We'll make sure you understand every bit of it, no sweat!
Problem 1: Unraveling the Rectangle's Length
Alright guys, let's tackle our first problem. We're dealing with a rectangle, and we know its area is a whopping 6x² + 7x - 3. We're also given that its width is a neat (2x + 3). Your mission, should you choose to accept it, is to find the length of this rectangle. This is where our algebraic skills come into play, specifically with factoring quadratic expressions. Remember, the area of a rectangle is simply its length multiplied by its width (Area = Length × Width). So, if we have the area and the width, finding the length means we need to divide the area by the width. In other words, we need to factor the quadratic expression 6x² + 7x - 3 and see which of the factors matches the options provided. Let's break down the factoring process. We're looking for two binomials that, when multiplied together, give us 6x² + 7x - 3. This is a classic case of factoring a trinomial of the form ax² + bx + c. We need to find two numbers that multiply to a*c (which is 6 * -3 = -18) and add up to b (which is 7). Let's list the pairs of factors for -18: (1, -18), (-1, 18), (2, -9), (-2, 9), (3, -6), (-3, 6). Now, let's see which pair adds up to 7. Bingo! (-2, 9) is our pair because -2 + 9 = 7. Now we can rewrite the middle term (7x) using these two numbers: 6x² - 2x + 9x - 3. The next step is to factor by grouping. Let's group the first two terms and the last two terms: (6x² - 2x) + (9x - 3). Factor out the greatest common factor (GCF) from each group. From the first group, the GCF is 2x, leaving us with 2x(3x - 1). From the second group, the GCF is 3, leaving us with 3(3x - 1). Notice that we now have a common binomial factor, (3x - 1). So, we can factor out (3x - 1) from the entire expression: (3x - 1)(2x + 3). Now, compare this to the given information: Area = (3x - 1)(2x + 3) and Width = (2x + 3). By simple comparison, we can see that the length of the rectangle must be (3x - 1). This matches option (a)! How cool is that? We used factoring to solve a geometry problem. It's all connected in the world of math, guys!
Problem 2: Identifying the Prime Number
Moving on to our next challenge, we're diving into the fascinating world of prime numbers. The question is: Which of the following is a prime number? We're given four options: (a) 115, (b) 119, (c) 127, and (d) 147. Remember, a prime number is a whole number greater than 1 that has only two divisors: 1 and itself. Any number that has more than two divisors is called a composite number. So, our task is to test each of these numbers to see if they have any divisors other than 1 and themselves. Let's start with option (a), 115. Numbers ending in 5 are always divisible by 5. So, 115 ÷ 5 = 23. Since 115 has divisors 5 and 23 (besides 1 and 115), it's a composite number. Moving on to option (b), 119. Let's try dividing by small prime numbers. Is it divisible by 2? No, it's odd. Is it divisible by 3? The sum of its digits is 1+1+9 = 11, which is not divisible by 3, so 119 is not divisible by 3. Is it divisible by 5? No, it doesn't end in 0 or 5. How about 7? Let's try: 119 ÷ 7. Well, 7 x 10 = 70, and 119 - 70 = 49. And we know 7 x 7 = 49. So, 119 ÷ 7 = 10 + 7 = 17. Aha! 119 has divisors 7 and 17, so it's also a composite number. Now, let's check option (c), 127. We need to test for divisibility by prime numbers up to the square root of 127. The square root of 127 is approximately 11.2. So, we need to test primes 2, 3, 5, 7, and 11. We already know it's not divisible by 2 (it's odd) or 5 (doesn't end in 0 or 5). Sum of digits 1+2+7 = 10, not divisible by 3. Let's try 7: 127 ÷ 7. 7 x 10 = 70, 127 - 70 = 57. 7 x 8 = 56. So, 127 is not divisible by 7 (remainder is 1). Let's try 11: 127 ÷ 11. 11 x 10 = 110, 127 - 110 = 17. 11 x 1 = 11. Not divisible by 11 either (remainder is 6). Since 127 is not divisible by any prime number less than or equal to its square root, 127 is a prime number! Awesome! Just for completeness, let's quickly check option (d), 147. Sum of digits 1+4+7 = 12, which is divisible by 3. So, 147 is divisible by 3. 147 ÷ 3 = 49. So, 147 is composite. Therefore, the only prime number in the list is 127.
Problem 3: Exploring the Cube Identity
Finally, let's dive into our third problem, which involves a neat algebraic identity. We're given the condition: a/b + b/a = 1. And the question is to find the value of a³ + b³. This problem is all about manipulating algebraic expressions and recognizing useful patterns. First, let's simplify the given condition a/b + b/a = 1. To combine the fractions on the left side, we need a common denominator, which is ab. So, we get (a² + b²)/ab = 1. Now, multiply both sides by ab to get rid of the denominator: a² + b² = ab. Our goal is to find a³ + b³. We know a very useful algebraic identity for the sum of cubes: a³ + b³ = (a + b)(a² - ab + b²). We also have another identity that might be helpful: (a + b)² = a² + 2ab + b². Let's rearrange our simplified condition a² + b² = ab. If we subtract ab from both sides, we get a² - ab + b² = 0. Look closely at this expression! It's exactly the second factor in our sum of cubes identity a³ + b³ = (a + b)(a² - ab + b²). Since we found that a² - ab + b² = 0, we can substitute this into the sum of cubes identity: a³ + b³ = (a + b)(0). Anything multiplied by zero is zero. Therefore, a³ + b³ = 0. This is a really elegant result that comes from a simple starting point. It's a great example of how algebraic identities can simplify complex expressions. Remember these tricks, guys, they'll save you a lot of time and effort in future math problems!
Conclusion:
There you have it, team! We've conquered problems involving rectangle area, prime number identification, and algebraic cube identities. Each problem tested different aspects of our mathematical knowledge, from factoring and divisibility rules to algebraic manipulation. Keep practicing these types of problems, and you'll become a math superstar in no time. Remember, math is not just about formulas; it's about logical thinking and problem-solving. So, keep those brains sharp and keep exploring the wonderful world of numbers!