Solving $2x^3 - X^2 - 98x + 49 = 0$: A Step-by-Step Guide

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Welcome, math enthusiasts! Today, we're diving deep into the fascinating world of algebra to tackle a specific cubic equation: $2x^3 - x^2 - 98x + 49 = 0$. Cubic equations, with their characteristic $x^3$ term, can sometimes seem intimidating. However, like any mathematical puzzle, they can be solved with the right strategies and a clear understanding of the underlying principles. Our goal here is not just to find the roots of this particular equation but to equip you with the knowledge to approach similar problems with confidence. We'll break down the process into manageable steps, making it accessible even if cubic equations aren't your everyday bread and butter. So, let's roll up our sleeves and get ready to unravel the mysteries hidden within this polynomial!

Understanding Cubic Equations

Before we plunge into solving our specific equation, let's take a moment to appreciate what a cubic equation is. In its most general form, a cubic equation is a polynomial equation of the third degree, meaning it can be written as $ax^3 + bx^2 + cx + d = 0$, where $a$, $b$, $c$, and $d$ are coefficients, and importantly, $a \neq 0$. The presence of the $x^3$ term is what defines it as cubic. A cubic equation will always have exactly three roots, although these roots might be real or complex, and they might be distinct or repeated. The Fundamental Theorem of Algebra guarantees this. Solving these equations can range from straightforward to quite complex, depending on the coefficients and the nature of the roots. Some cubic equations can be solved by factoring, while others might require more advanced techniques like the Rational Root Theorem, synthetic division, or even numerical methods. The equation we're focusing on today, $2x^3 - x^2 - 98x + 49 = 0$, fits into the category that can be tackled effectively with some clever algebraic manipulation.

Strategy: Factoring by Grouping

For the cubic equation $2x^3 - x^2 - 98x + 49 = 0$, one of the most efficient methods to try first is factoring by grouping. This technique is particularly useful when a polynomial has four terms, just like ours. The core idea is to group the terms into pairs, factor out the greatest common factor (GCF) from each pair, and then see if a common binomial factor emerges. If it does, we can factor that binomial out, leaving us with a simpler expression, often a quadratic, which we can then solve. Let's apply this to our equation. We'll group the first two terms and the last two terms:

$$(2x^3 - x^2) + (-98x + 49) = 0$$

Now, let's factor out the GCF from each group. In the first group, $(2x^3 - x^2)$, the GCF is $x^2$. Factoring this out gives us $x^2(2x - 1)$. In the second group, $(-98x + 49)$, the GCF is $-49$. Factoring this out gives us $-49(2x - 1)$. Notice that by factoring out $-49$ (instead of just $49$), we've managed to make the binomial factor inside the parentheses the same as in the first group. This is crucial for factoring by grouping to work!

So, our equation now looks like this:

$$x^2(2x - 1) - 49(2x - 1) = 0$$

As you can see, we have a common binomial factor of $(2x - 1)$. We can now factor this out from the entire expression:

$$(2x - 1)(x^2 - 49) = 0$$

We have successfully factored our cubic equation into the product of a linear term and a quadratic term. This is a significant step towards finding the roots.

Solving the Factored Equation

With our equation factored into $(2x - 1)(x^2 - 49) = 0$, we can now use the zero product property. This property states that if the product of two or more factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for $x$:

Factor 1: $2x - 1 = 0$

To solve for $x$, we add 1 to both sides:

$$2x = 1$$

Then, we divide both sides by 2:

$$x = \frac{1}{2}$$

This is our first root.

Factor 2: $x^2 - 49 = 0$

This is a quadratic equation. We can solve it by recognizing it as a difference of squares, $a^2 - b^2 = (a - b)(a + b)$. Here, $a = x$ and $b = 7$. So, we can rewrite $x^2 - 49$ as $(x - 7)(x + 7)$.

Setting this factored form to zero gives us:

$$(x - 7)(x + 7) = 0$$

Now, we apply the zero product property again:

• $x - 7 = 0 \implies x = 7$
• $x + 7 = 0 \implies x = -7$

These are our second and third roots.

Verifying the Roots

To ensure our calculations are correct, it's always a good practice to verify the roots by substituting them back into the original equation $2x^3 - x^2 - 98x + 49 = 0$. Let's check each root:

1. For $x = \frac{1}{2}$:

$$2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 - 98\left(\frac{1}{2}\right) + 49$$

$$= 2\left(\frac{1}{8}\right) - \frac{1}{4} - 49 + 49$$

$$= \frac{1}{4} - \frac{1}{4} - 49 + 49 = 0$$

This root is correct.

2. For $x = 7$:

$$2(7)^3 - (7)^2 - 98(7) + 49$$

$$= 2(343) - 49 - 686 + 49$$

$$= 686 - 49 - 686 + 49 = 0$$

This root is also correct.

3. For $x = -7$:

$$2(-7)^3 - (-7)^2 - 98(-7) + 49$$

$$= 2(-343) - 49 + 686 + 49$$

$$= -686 - 49 + 686 + 49 = 0$$

This root is correct as well.

All three roots satisfy the original equation, confirming our solutions.

Alternative Methods: Rational Root Theorem and Synthetic Division

While factoring by grouping was very effective for $2x^3 - x^2 - 98x + 49 = 0$, it's beneficial to know other approaches, especially when factoring by grouping doesn't immediately present itself. The Rational Root Theorem is a powerful tool for finding potential rational roots of a polynomial. For a polynomial $a_nx^n + ... + a_0 = 0$, any rational root must be of the form $p/q$, where $p$ is a factor of the constant term ($a_0$) and $q$ is a factor of the leading coefficient ($a_n$).

In our equation, $2x^3 - x^2 - 98x + 49 = 0$, the constant term is $49$ and the leading coefficient is $2$.

Factors of $49$ ($p$): $\pm 1, \pm 7, \pm 49$ Factors of $2$ ($q$): $\pm 1, \pm 2$

Possible rational roots ($p/q$): $\pm 1, \pm 7, \pm 49, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{49}{2}$.

We could then test these values using synthetic division. Synthetic division is a shorthand method for dividing a polynomial by a linear factor of the form $(x - c)$. If the remainder is zero, then $c$ is a root of the polynomial. For instance, if we test $x = \frac{1}{2}$ using synthetic division:

1/2 | 2  -1  -98   49
    |    1    0  -49
    -----------------
      2   0  -98    0

The remainder is $0$, confirming that $x = \frac{1}{2}$ is a root. The resulting coefficients ($2, 0, -98$) represent the depressed polynomial, which is $2x^2 + 0x - 98 = 2x^2 - 98$. Setting this to zero, $2x^2 - 98 = 0$, we get $2x^2 = 98$, $x^2 = 49$, and thus $x = \pm 7$. This method also yields the same roots, showcasing its utility.

Conclusion

We've successfully solved the cubic equation $2x^3 - x^2 - 98x + 49 = 0$ using the powerful technique of factoring by grouping. We found the roots to be $x = \frac{1}{2}$, $x = 7$, and $x = -7$. We also touched upon alternative methods like the Rational Root Theorem and synthetic division, which are invaluable when factoring by grouping isn't straightforward. Understanding these different approaches enhances your problem-solving toolkit in algebra. The journey through solving polynomial equations is a fundamental aspect of mathematics, opening doors to more complex concepts in calculus, differential equations, and beyond. Keep practicing, and don't shy away from those challenging equations – they are opportunities for growth!

For further exploration into algebraic equations and polynomial theory, you can refer to resources like Khan Academy's Algebra section or Wolfram MathWorld's Polynomial page.