Factoring Polynomials Find The Factored Form Of 27y^3 + 125

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Hey guys! Let's dive into factoring polynomials, specifically the sum of cubes. We're going to break down the polynomial $27y^3 + 125$ step-by-step, so you can confidently choose the correct factored form. Factoring polynomials can seem daunting at first, but with the right approach and understanding of key formulas, it becomes a manageable and even enjoyable task. This guide will not only provide the solution but also explain the underlying principles and common pitfalls to avoid. So, buckle up, and let's get started on this mathematical journey!

Understanding the Sum of Cubes Formula

Before we tackle the specific problem, it's crucial to understand the sum of cubes formula. This formula is our main tool for factoring expressions in the form of $a^3 + b^3$. The formula states:

a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

This formula might look a bit intimidating at first glance, but it's actually quite straightforward once you break it down. The left side of the equation represents the sum of two terms, each raised to the power of three (cubed). The right side of the equation provides the factored form, which consists of two factors:

  1. A binomial factor: $(a + b)$, which is the sum of the cube roots of the original terms.
  2. A trinomial factor: $(a^2 - ab + b^2)$, which is derived from the binomial factor. It consists of the square of the first term, minus the product of the two terms, plus the square of the second term.

Understanding the pattern and the components of this formula is key to successfully factoring sums of cubes. We'll apply this formula to our specific polynomial, $27y^3 + 125$, in the following sections. Let's make sure we've got this formula down solid – it's the foundation for what we're about to do!

Identifying $a$ and $b$

The first step in applying the sum of cubes formula to our polynomial, $27y^3 + 125$, is to identify the values of $a$ and $b$. Remember, we need to express each term in the polynomial as a perfect cube. This means finding the cube root of each term.

Let's start with the first term, $27y^3$. We need to find a term that, when cubed, gives us $27y^3$. We can break this down into two parts: the coefficient (27) and the variable part ($y^3$).

The cube root of 27 is 3, since $3^3 = 3 \times 3 \times 3 = 27$. The cube root of $y^3$ is simply $y$, since $(y)^3 = y \times y \times y = y^3$. Therefore, the cube root of $27y^3$ is $3y$. So, we can say that $a = 3y$.

Now let's move on to the second term, 125. We need to find a number that, when cubed, equals 125. You might recognize that 125 is a perfect cube, and its cube root is 5, since $5^3 = 5 \times 5 \times 5 = 125$. Thus, we have $b = 5$.

So, to recap, we've identified:

  • a=3ya = 3y

  • b=5b = 5

These values are crucial for plugging into the sum of cubes formula. With these values in hand, we're ready to move on to the next step: applying the formula to find the factored form. Trust me, guys, once you get the hang of identifying $a$ and $b$, the rest is smooth sailing!

Applying the Sum of Cubes Formula

Now that we've identified $a = 3y$ and $b = 5$, we can plug these values into the sum of cubes formula:

a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

Substituting our values, we get:

(3y)3+(5)3=(3y+5)((3y)2−(3y)(5)+(5)2)(3y)^3 + (5)^3 = (3y + 5)((3y)^2 - (3y)(5) + (5)^2)

Now, let's simplify the expression on the right side. We need to calculate the square of $3y$, the product of $3y$ and 5, and the square of 5.

  • (3y)2=(3y)×(3y)=9y2(3y)^2 = (3y) \times (3y) = 9y^2

  • (3y)(5)=15y(3y)(5) = 15y

  • (5)2=5×5=25(5)^2 = 5 \times 5 = 25

Plugging these simplified terms back into our equation, we get:

(3y+5)(9y2−15y+25)(3y + 5)(9y^2 - 15y + 25)

This is the factored form of the polynomial $27y^3 + 125$. We've successfully applied the sum of cubes formula and simplified the resulting expression. It's like solving a puzzle, piece by piece, and seeing the final picture come together. You're doing great! Now, let's compare our result with the given choices to select the correct answer.

Comparing with the Choices

Alright, we've factored the polynomial $27y^3 + 125$ and arrived at the factored form: $(3y + 5)(9y^2 - 15y + 25)$. Now, let's compare this result with the choices provided in the original question:

a. $(3y + 5)(9y + 25)$ b. $(3y + 5)(9y^2 - 15y + 25)$ c. Not factorable d. $(3y - 5)(9y^2 + 15y + 25)$

By comparing our factored form with the given choices, we can clearly see that option b matches our result exactly. The factor $(3y + 5)$ is present in both our solution and option b, and the same goes for the trinomial factor $(9y^2 - 15y + 25)$.

Therefore, the correct answer is:

b. $(3y + 5)(9y^2 - 15y + 25)$

Options a, c, and d are incorrect. Option a has an incorrect trinomial factor, option c is incorrect because the polynomial is factorable, and option d has an incorrect sign in the binomial factor and the trinomial factor.

So, there you have it! We've successfully factored the sum of cubes polynomial and identified the correct factored form from the given choices. It's all about understanding the formula, applying it step-by-step, and carefully comparing the result. You're becoming factoring pros, guys!

Common Mistakes to Avoid

When factoring the sum of cubes, there are some common mistakes that students often make. Being aware of these pitfalls can help you avoid them and ensure you get the correct answer every time. Let's highlight some key areas to watch out for:

  1. Incorrect Signs in the Trinomial Factor: This is perhaps the most common mistake. Remember, the sum of cubes formula is:

    a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

    Notice the minus sign in front of the $ab$ term in the trinomial factor. It's easy to forget this and write a plus sign instead. Similarly, when factoring the difference of cubes ($a^3 - b^3$), the formula is:

    a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

    Here, the plus sign in front of the $ab$ term is crucial. Always double-check your signs to avoid this error.
  2. Forgetting to Square the Terms in the Trinomial Factor: The trinomial factor includes the terms $a^2$ and $b^2$. Make sure you actually square the terms $a$ and $b$ when constructing this factor. For example, if $a = 3y$, then $a^2$ should be $(3y)^2 = 9y^2$, not just $3y^2$.
  3. Incorrectly Identifying $a$ and $b$: As we discussed earlier, correctly identifying $a$ and $b$ is the foundation of factoring the sum or difference of cubes. Make sure you take the cube root of each term, not just the square root. For example, if you have $27y^3$, remember that $a = 3y$, not just $3$ or $y$.
  4. Thinking it's Always Factorable: While many sums and differences of cubes are factorable, not all polynomials are. However, if you can express the polynomial in the form $a^3 + b^3$ or $a^3 - b^3$, then it is factorable using the respective formulas. If you can't identify perfect cubes, then it might not be factorable using these specific methods.
  5. Skipping Steps: It's tempting to rush through the factoring process, especially once you feel comfortable with the formula. However, skipping steps can lead to careless errors. Take your time, write out each step clearly, and double-check your work. It's better to be thorough than to make a mistake that could have been easily avoided.

By keeping these common mistakes in mind, you can significantly improve your accuracy and confidence in factoring the sum and difference of cubes. Remember, practice makes perfect, so keep working at it!

Practice Problems

To really solidify your understanding of factoring the sum of cubes, let's try a few practice problems. Working through these examples will help you become more comfortable with the formula and the steps involved. Plus, it's a great way to identify any areas where you might need a little extra practice. So, grab a pencil and paper, and let's get to it!

Practice Problem 1:

Factor the polynomial $8x^3 + 1$.

Solution:

  1. Identify $a$ and $b$:

    • a=8x33=2xa = \sqrt[3]{8x^3} = 2x

    • b=13=1b = \sqrt[3]{1} = 1

  2. Apply the sum of cubes formula:

    a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

    Substitute $a = 2x$ and $b = 1$:

    (2x)3+(1)3=(2x+1)((2x)2−(2x)(1)+(1)2)(2x)^3 + (1)^3 = (2x + 1)((2x)^2 - (2x)(1) + (1)^2)

  3. Simplify:

    (2x+1)(4x2−2x+1)(2x + 1)(4x^2 - 2x + 1)

So, the factored form of $8x^3 + 1$ is $(2x + 1)(4x^2 - 2x + 1)$.

Practice Problem 2:

Factor the polynomial $64y^3 + 27$.

Solution:

  1. Identify $a$ and $b$:

    • a=64y33=4ya = \sqrt[3]{64y^3} = 4y

    • b=273=3b = \sqrt[3]{27} = 3

  2. Apply the sum of cubes formula:

    a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

    Substitute $a = 4y$ and $b = 3$:

    (4y)3+(3)3=(4y+3)((4y)2−(4y)(3)+(3)2)(4y)^3 + (3)^3 = (4y + 3)((4y)^2 - (4y)(3) + (3)^2)

  3. Simplify:

    (4y+3)(16y2−12y+9)(4y + 3)(16y^2 - 12y + 9)

Thus, the factored form of $64y^3 + 27$ is $(4y + 3)(16y^2 - 12y + 9)$.

Practice Problem 3:

Factor the polynomial $125z^3 + 8$.

Solution:

  1. Identify $a$ and $b$:

    • a=125z33=5za = \sqrt[3]{125z^3} = 5z

    • b=83=2b = \sqrt[3]{8} = 2

  2. Apply the sum of cubes formula:

    a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

    Substitute $a = 5z$ and $b = 2$:

    (5z)3+(2)3=(5z+2)((5z)2−(5z)(2)+(2)2)(5z)^3 + (2)^3 = (5z + 2)((5z)^2 - (5z)(2) + (2)^2)

  3. Simplify:

    (5z+2)(25z2−10z+4)(5z + 2)(25z^2 - 10z + 4)

Therefore, the factored form of $125z^3 + 8$ is $(5z + 2)(25z^2 - 10z + 4)$.

By working through these practice problems, you've gained valuable experience in applying the sum of cubes formula. Remember, the key is to correctly identify $a$ and $b$, plug them into the formula, and simplify. Keep practicing, and you'll become a factoring master in no time!

Conclusion

We've reached the end of our journey into factoring the sum of cubes, and what a journey it has been! We started by understanding the sum of cubes formula, then applied it step-by-step to the polynomial $27y^3 + 125$. We learned how to identify $a$ and $b$, substitute them into the formula, and simplify the resulting expression. We then compared our factored form with the given choices and confidently selected the correct answer: $(3y + 5)(9y^2 - 15y + 25)$.

But we didn't stop there! We also explored common mistakes to avoid, ensuring you're well-equipped to tackle any sum of cubes factoring problem that comes your way. And, to really drive the concepts home, we worked through several practice problems, solidifying your understanding and boosting your confidence.

Factoring polynomials, especially the sum and difference of cubes, is a fundamental skill in algebra. Mastering this skill opens doors to more advanced topics and problem-solving techniques. So, whether you're a student preparing for an exam or simply someone who enjoys the challenge of mathematical puzzles, understanding factoring is a valuable asset.

Remember, guys, the key to success in mathematics is practice, practice, practice! The more you work with these formulas and techniques, the more natural they will become. Don't be afraid to make mistakes – they're a crucial part of the learning process. And most importantly, have fun with it! Math can be challenging, but it can also be incredibly rewarding.

So, go forth and conquer those polynomials! You've got the tools, the knowledge, and the confidence to succeed. Keep practicing, keep learning, and keep exploring the fascinating world of mathematics. Until next time, happy factoring!