Factoring Quadratics: Finding The True Statement

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Hey everyone! Let's dive into the world of factoring quadratic expressions. We're going to break down the expression βˆ’2h2βˆ’15hβˆ’7-2h^2 - 15h - 7 and figure out which of the provided statements is correct. This is a common type of problem in algebra, so understanding the process is super helpful. We'll examine each option and see if it holds water. Ready to get started? Let's go!

Understanding Quadratic Expressions and Factoring

Alright, before we jump into the options, let's quickly recap what quadratic expressions and factoring are all about. A quadratic expression is an algebraic expression of the form ax2+bx+cax^2 + bx + c, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. In our case, we have βˆ’2h2βˆ’15hβˆ’7-2h^2 - 15h - 7. The process of factoring involves breaking down a quadratic expression into a product of simpler expressions, usually two binomials. Think of it like this: factoring is the reverse of expanding (or multiplying out) binomials. When we factor a quadratic, we're essentially looking for two binomials that, when multiplied together, give us the original quadratic expression. The goal of this question is to determine the correct factors of the given quadratic expression.

To make this super clear, let's look at an example. If we had x2+5x+6x^2 + 5x + 6, we could factor it into (x+2)(x+3)(x + 2)(x + 3). If you multiply (x+2)(x + 2) and (x+3)(x + 3), you'll get back x2+5x+6x^2 + 5x + 6. So, the factors of the quadratic are (x+2)(x + 2) and (x+3)(x + 3). This concept of factoring is fundamental in algebra and is essential for solving quadratic equations and simplifying expressions. The best approach to tackling such problems is through systematic checking of factors, and let's go over how we can do that for the given question. Let's get our hands dirty and determine the true statement. We'll be using different approaches to do that like using the factor theorem. The factor theorem helps to test a given factor by substituting the value of 'h' that makes the potential factor equal to zero in the quadratic expression. If the result is zero, then the potential factor is a correct factor. This method helps to test our options and find the correct answer.

Now, let's use all that we have learned to solve our problem. Remember the quadratic equation we need to solve: βˆ’2h2βˆ’15hβˆ’7-2h^2 - 15h - 7.

Analyzing the Answer Choices

Now, let's go through the answer choices one by one. We'll use a combination of techniques, like trying to factor the expression, and using the factor theorem to verify the correctness of each option. This way, we will eliminate wrong answers and find the correct one.

A. One of the factors is (h+2)(h + 2).

Let's test this option. If (h+2)(h + 2) is a factor, then the value of 'h' that makes the factor equal to zero is h=βˆ’2h = -2. Substitute this value into the quadratic expression: βˆ’2(βˆ’2)2βˆ’15(βˆ’2)βˆ’7-2(-2)^2 - 15(-2) - 7. This simplifies to βˆ’2(4)+30βˆ’7=βˆ’8+30βˆ’7=15-2(4) + 30 - 7 = -8 + 30 - 7 = 15. Since the result is not zero, (h+2)(h + 2) is not a factor of the quadratic expression. So, option A is incorrect. We've eliminated the first option, and we're moving forward. Keep up the good work; we are almost there!

B. One of the factors is (3hβˆ’2)(3h - 2).

Now, we will test the second option. If (3hβˆ’2)(3h - 2) is a factor, then the value of 'h' that makes the factor equal to zero is h = rac{2}{3}. Substitute this into the quadratic expression: -2( rac{2}{3})^2 - 15( rac{2}{3}) - 7. This simplifies to -2( rac{4}{9}) - 10 - 7 = - rac{8}{9} - 17. The result is not zero. Therefore, (3hβˆ’2)(3h - 2) is not a factor of the quadratic expression. Option B is also incorrect, so let's move forward and analyze other options to find the correct answer.

C. One of the factors is (2h+1)(2h + 1).

Let's check option C. If (2h+1)(2h + 1) is a factor, then the value of 'h' that makes the factor equal to zero is h = - rac{1}{2}. Substitute this into the quadratic expression: -2(- rac{1}{2})^2 - 15(- rac{1}{2}) - 7. This simplifies to -2( rac{1}{4}) + rac{15}{2} - 7 = - rac{1}{2} + rac{15}{2} - rac{14}{2} = 0. Since the result is zero, (2h+1)(2h + 1) is indeed a factor of the quadratic expression. Option C seems to be the correct answer. But let's verify the last option as well, just to be sure that we have the correct answer.

D. One of the factors is (hβˆ’7)(h - 7).

Let's check the last option, just to be certain. If (hβˆ’7)(h - 7) is a factor, then the value of 'h' that makes the factor equal to zero is h=7h = 7. Substitute this into the quadratic expression: βˆ’2(7)2βˆ’15(7)βˆ’7-2(7)^2 - 15(7) - 7. This simplifies to βˆ’2(49)βˆ’105βˆ’7=βˆ’98βˆ’105βˆ’7=βˆ’210-2(49) - 105 - 7 = -98 - 105 - 7 = -210. Since the result is not zero, (hβˆ’7)(h - 7) is not a factor. So, option D is incorrect.

The Correct Answer

After analyzing all the options, we can confidently say that the correct answer is C. One of the factors is (2h+1)(2h + 1). We found that when we substituted the value of 'h' which makes the factor zero into the quadratic expression, the result was also zero. This confirms that (2h+1)(2h + 1) is a factor of the original quadratic equation. Understanding this process will help you solve similar problems in the future. Always remember to break down the problem into smaller parts and verify the results, and you are good to go! Great job, everyone!