Factoring Polynomials Step-by-Step G(x) = X³ - 7x² + 8x - 2

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Hey guys! Let's dive into this cool math problem where we need to factor a polynomial. We have a polynomial $g(x) = x^3 - 7x^2 + 8x - 2$, and we know that 1 is a zero of this polynomial. This means that if we plug in $x = 1$ into $g(x)$, we get zero. This is super helpful because it tells us that $(x - 1)$ is a factor of $g(x)$. Now, our mission is to express $g(x)$ as a product of linear factors. Buckle up, because we’re going on a mathematical adventure!

Using Synthetic Division to Simplify the Polynomial

First off, let’s use synthetic division. Synthetic division is a neat way to divide a polynomial by a linear factor. It's much quicker than long division, especially when we know a zero of the polynomial. Here’s how we set it up:

Write down the coefficients of $g(x)$: 1, -7, 8, and -2. Since 1 is a zero, we’ll use 1 as our divisor. Set up the synthetic division table like this:

1 | 1 -7 8 -2
 |______________

Now, let's perform the synthetic division:

  1. Bring down the first coefficient (1) to the bottom row.
1 | 1 -7 8 -2
 | 1
 |______________
 1
  1. Multiply the divisor (1) by the number we just brought down (1), and write the result under the next coefficient (-7).
1 | 1 -7 8 -2
 | 1
 |______________
 1

1 | 1 -7 8 -2
 | 1
 |______________
 1
  1. Add -7 and 1 to get -6, and write this under the line.
1 | 1 -7 8 -2
 | 1 -6
 |______________
 1 -6
  1. Multiply the divisor (1) by -6, and write the result under the next coefficient (8).
1 | 1 -7 8 -2
 | 1 -6
 |______________
 1 -6

1 | 1 -7 8 -2
 | 1 -6
 |______________
 1 -6
  1. Add 8 and -6 to get 2, and write this under the line.
1 | 1 -7 8 -2
 | 1 -6 2
 |______________
 1 -6 2
  1. Multiply the divisor (1) by 2, and write the result under the last coefficient (-2).
1 | 1 -7 8 -2
 | 1 -6 2
 |______________
 1 -6 2

1 | 1 -7 8 -2
 | 1 -6 2
 |______________
 1 -6 2
  1. Add -2 and 2 to get 0. This is our remainder.
1 | 1 -7 8 -2
 | 1 -6 2
 |______________
 1 -6 2 0

So, the bottom row gives us the coefficients of the quotient polynomial. We started with a cubic polynomial ($x^3$), and since we divided by a linear factor, we're left with a quadratic polynomial. The coefficients 1, -6, and 2 correspond to $x^2 - 6x + 2$. Therefore, we can write $g(x)$ as:

g(x)=(x−1)(x2−6x+2)g(x) = (x - 1)(x^2 - 6x + 2)

Solving the Quadratic Equation

Now, we need to factor the quadratic $x^2 - 6x + 2$. Since it doesn't factor nicely using integers, we'll use the quadratic formula. The quadratic formula is a powerful tool that helps us find the roots of any quadratic equation in the form $ax^2 + bx + c = 0$. The formula is:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, $a = 1$, $b = -6$, and $c = 2$. Plugging these values into the quadratic formula, we get:

x=−(−6)±(−6)2−4(1)(2)2(1)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(2)}}{2(1)}

x=6±36−82x = \frac{6 \pm \sqrt{36 - 8}}{2}

x=6±282x = \frac{6 \pm \sqrt{28}}{2}

x=6±272x = \frac{6 \pm 2\sqrt{7}}{2}

x=3±7x = 3 \pm \sqrt{7}

So, the roots of the quadratic $x^2 - 6x + 2$ are $3 + \sqrt{7}$ and $3 - \sqrt{7}$.

Expressing g(x) as a Product of Linear Factors

Now that we have the roots of the quadratic, we can write it as a product of linear factors. Remember, if $r$ is a root of a polynomial, then $(x - r)$ is a factor. Therefore, the factors corresponding to the roots $3 + \sqrt{7}$ and $3 - \sqrt{7}$ are:

(x - (3 + \sqrt{7}))$ and $(x - (3 - \sqrt{7}))

So, we can write the quadratic as:

x2−6x+2=(x−(3+7))(x−(3−7))x^2 - 6x + 2 = (x - (3 + \sqrt{7}))(x - (3 - \sqrt{7}))

Finally, we can express $g(x)$ as a product of linear factors by combining the factor $(x - 1)$ with the factored quadratic:

g(x)=(x−1)(x−(3+7))(x−(3−7))g(x) = (x - 1)(x - (3 + \sqrt{7}))(x - (3 - \sqrt{7}))

Final Answer: Factoring g(x) Completely

And there we have it! We've successfully expressed $g(x)$ as a product of linear factors. To recap, we used synthetic division to reduce the cubic polynomial to a quadratic, then we used the quadratic formula to find the roots of the quadratic, and finally, we wrote $g(x)$ as a product of linear factors:

g(x)=(x−1)(x−(3+7))(x−(3−7))g(x) = (x - 1)(x - (3 + \sqrt{7}))(x - (3 - \sqrt{7}))

Isn't math cool? We took a seemingly complex polynomial and broke it down into its simplest components. Remember, practice makes perfect, so keep tackling those math problems!

Understanding the Zeros of a Polynomial

Let's zoom in a bit on why finding the zeros of a polynomial is such a big deal. The zeros of a polynomial, also known as roots, are the values of $x$ that make the polynomial equal to zero. These zeros tell us a lot about the behavior of the polynomial function, especially its graph. For instance, each real zero corresponds to an x-intercept on the graph of the polynomial. In our case, the zeros of $g(x)$ are 1, $3 + \sqrt{7}$, and $3 - \sqrt{7}$. This means that the graph of $g(x)$ crosses the x-axis at these three points.

Knowing the zeros helps us sketch the graph of the polynomial, understand its end behavior, and solve related equations and inequalities. The Fundamental Theorem of Algebra tells us that a polynomial of degree $n$ has exactly $n$ complex roots, counting multiplicity. This is why factoring polynomials into linear factors is so crucial – it directly reveals these roots. When we express a polynomial as a product of linear factors, each factor corresponds to a root, making it easier to analyze and solve.

In the context of $g(x) = x^3 - 7x^2 + 8x - 2$, knowing that 1 is a zero was our starting point. It allowed us to use synthetic division, which simplified the problem significantly. Without knowing an initial zero, factoring a cubic polynomial can be quite challenging. The rational root theorem can help us find potential rational roots, but in this case, synthetic division with the given zero made the process much more straightforward. Once we had the quadratic factor, using the quadratic formula was the most direct way to find the remaining roots, as the quadratic did not factor easily by inspection.

Practical Applications of Polynomial Factorization

Polynomial factorization isn't just an abstract mathematical exercise; it has practical applications in various fields. For example, in engineering, polynomials are used to model various systems, and their zeros can represent critical values or stable states. In physics, polynomial equations can describe the motion of objects, and finding the roots can help determine when an object reaches a certain position. In computer graphics, polynomials are used to create curves and surfaces, and understanding their roots is important for rendering and manipulation.

Moreover, in cryptography, polynomials play a crucial role in creating secure communication systems. The difficulty of factoring large polynomials is used as the basis for certain encryption methods. By breaking down complex polynomials into simpler factors, we can solve problems in diverse areas, from designing bridges to securing digital communications.

Understanding polynomial factorization is also essential for more advanced mathematical concepts. In calculus, it helps in finding limits, derivatives, and integrals of rational functions. In linear algebra, it is used to find eigenvalues and eigenvectors of matrices, which are crucial in solving systems of differential equations and analyzing linear transformations. So, the skills we've honed in this problem are not just for academic purposes; they're valuable tools for a wide range of applications.

Tips for Mastering Polynomial Factorization

So, how can you become a polynomial factorization pro? Here are a few tips and tricks:

  1. Know Your Special Cases: Recognizing patterns like the difference of squares $(a^2 - b^2 = (a - b)(a + b))$ and perfect square trinomials $(a^2 \pm 2ab + b^2 = (a \pm b)^2)$ can save you a lot of time.

  2. Use Synthetic Division: Synthetic division is a game-changer when you know a zero of the polynomial. It simplifies the polynomial and makes it easier to find the remaining factors.

  3. Master the Quadratic Formula: The quadratic formula is your best friend for solving quadratic equations that don't factor easily. Make sure you know it by heart and can apply it correctly.

  4. Look for Common Factors: Always check if there's a common factor you can pull out from all terms. This can simplify the polynomial and make it easier to factor.

  5. Practice, Practice, Practice: The more you practice, the better you'll get at recognizing patterns and applying the appropriate techniques. Work through a variety of problems to build your skills.

  6. Use the Rational Root Theorem: If you need to find potential rational roots, the Rational Root Theorem can help you narrow down the possibilities.

  7. Don't Give Up: Factoring polynomials can be challenging, but don't get discouraged. Keep trying different approaches, and you'll eventually find the solution.

Conclusion: The Art and Science of Polynomial Factorization

Factoring polynomials is both an art and a science. It requires a combination of knowledge, skill, and intuition. By understanding the underlying principles, mastering the techniques, and practicing regularly, you can become a proficient polynomial factorizer. The problem we tackled today, expressing $g(x) = x^3 - 7x^2 + 8x - 2$ as a product of linear factors, is a great example of how these skills come together.

We started with the knowledge that 1 is a zero, which allowed us to use synthetic division to simplify the polynomial. We then applied the quadratic formula to find the remaining roots and expressed the quadratic as a product of linear factors. Finally, we combined these factors to write $g(x)$ as a product of linear factors:

g(x)=(x−1)(x−(3+7))(x−(3−7))g(x) = (x - 1)(x - (3 + \sqrt{7}))(x - (3 - \sqrt{7}))

This process not only solves the problem but also deepens our understanding of polynomials, their roots, and their factors. So keep exploring, keep learning, and keep factoring! You've got this!