Solve For A, B, And C In Polynomial Identity

by Andrew McMorgan 45 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a rather intriguing polynomial identity. If you're a math whiz or just enjoy a good brain teaser, you're in for a treat! We're going to break down the equation (1-2 \sqrt{2}) x^3+(a-x)(3 x+b)+x(1-c)^{ rac{1}{2}} \equiv(1-a) x^3-3 x^2+x-2^{ rac{3}{2}}+3 and figure out the exact values of aa, bb, and cc that make this identity hold true. This isn't just about finding numbers; it's about understanding how polynomial identities work and the systematic way we can unravel them. So, grab your notebooks, maybe a calculator if you're feeling fancy, and let's get cracking on this problem. We'll go step-by-step, ensuring that even if you're not a seasoned mathematician, you can follow along and appreciate the elegance of algebraic manipulation. It's all about breaking down complex problems into manageable parts, and this identity is a perfect example of that. We'll be expanding terms, comparing coefficients, and using the fundamental principle that for two polynomials to be identical, all their corresponding coefficients must be equal. Let's get started on this mathematical adventure!

Unpacking the Polynomial Identity

Alright, let's start by taking a really good look at the equation we've been given: (1-2 \sqrt{2}) x^3+(a-x)(3 x+b)+x(1-c)^{ rac{1}{2}} \equiv(1-a) x^3-3 x^2+x-2^{\frac{3}{2}}+3. The symbol ≡\equiv here means that the expression on the left is identical to the expression on the right for all values of xx. This is the core concept we'll be using – polynomial identity. For two polynomials to be identical, every single one of their corresponding coefficients (the numbers multiplying each power of xx) must be exactly the same. It's like having two identical twins; if you look at every feature, they match up. So, our mission, should we choose to accept it, is to expand the left side of the equation until it's in a similar form to the right side, which is already nicely expanded. This will allow us to directly compare the coefficients of x3x^3, x2x^2, x1x^1, and the constant terms. This systematic comparison is the key to solving for our unknowns: aa, bb, and cc. We need to be super careful with our algebra here, especially with the terms involving square roots and potential negative signs. Each step needs to be precise. Let's break down the left side piece by piece. We have the first term (1−22)x3(1-2 \sqrt{2}) x^3, which is already in a good form. Next, we have (a−x)(3x+b)(a-x)(3 x+b). This is a product of two binomials, and we'll need to use the distributive property (or FOIL method, if you prefer) to expand it. Remember, (a−x)(3x+b)=a(3x+b)−x(3x+b)=3ax+ab−3x2−bx(a-x)(3x+b) = a(3x+b) - x(3x+b) = 3ax + ab - 3x^2 - bx. After expanding, we'll group like terms. The last part of the left side is x(1−c)12x(1-c)^{\frac{1}{2}}. This term is a bit simpler; it's just xx multiplied by a constant value (1−c)12(1-c)^{\frac{1}{2}}. Don't let the exponent 12\frac{1}{2} (which means a square root) throw you off; it's just a constant number. Once we have the fully expanded and simplified left side, we'll equate its coefficients to the corresponding coefficients on the right side. This will give us a system of equations that we can solve. Get ready, because it's going to be a wild ride through algebra!

Expanding and Simplifying the Left Side

Now, let's get our hands dirty with the actual expansion and simplification of the left-hand side of our polynomial identity. We have (1-2 \sqrt{2}) x^3+(a-x)(3 x+b)+x(1-c)^{ rac{1}{2}}. We've already identified the parts we need to work on. The first term, (1−22)x3(1-2 \sqrt{2}) x^3, is fine as it is. The real work begins with the middle term: (a−x)(3x+b)(a-x)(3x+b). Using the distributive property, we get: $ (a-x)(3x+b) = a(3x) + a(b) - x(3x) - x(b) $ $ = 3ax + ab - 3x^2 - bx $ Now, let's rearrange this to group terms by powers of xx: $ = -3x^2 + (3a - b)x + ab $ Fantastic! We've handled the binomial multiplication. Now, let's look at the third term: x(1−c)12x(1-c)^{\frac{1}{2}}. This is straightforward. Let K=(1−c)12K = (1-c)^{\frac{1}{2}}. Then the term is simply KxKx. So, the entire left side becomes: $ (1-2

$\sqrt{2}) x^3 + (-3x^2 + (3a - b)x + ab) + Kx $ Combining like terms for the left side, we get: $ (1-2

$\sqrt{2}) x^3 - 3x^2 + ((3a - b) + K)x + ab $ Let's substitute KK back in: $ (1-2

$\sqrt{2}) x^3 - 3x^2 + ((3a - b) + (1-c)^{\frac{1}{2}})x + ab $ Phew! That's the fully expanded and simplified left side. Now, we need to compare this to the right side of the identity, which is (1−a)x3−3x2+x−232+3(1-a) x^3-3 x^2+x-2^{\frac{3}{2}}+3. The right side is already in its standard polynomial form. The crucial part is realizing that for the identity to hold, the coefficient of each power of xx on the left must match the coefficient of the same power of xx on the right. This is the fundamental principle of polynomial identity that will allow us to set up our equations. We are so close to finding aa, bb, and cc. The next step is where the magic happens – equating these coefficients.

Equating Coefficients and Forming Equations

With the left side of our polynomial identity neatly expanded and simplified to $(1-2

2)x3−3x2+((3a−b)+(1−c)12)x+ab\sqrt{2}) x^3 - 3x^2 + ((3a - b) + (1-c)^{\frac{1}{2}})x + ab, and the right side given as (1−a)x3−3x2+x−232+3(1-a) x^3-3 x^2+x-2^{\frac{3}{2}}+3, we can now perform the critical step: equating the coefficients. Remember, for two polynomials to be identical for all values of xx, the coefficient of each corresponding power of xx must be equal. This is the bedrock of solving such problems, guys. Let's break it down by the powers of xx:

1. Coefficients of x3x^3: On the left side, the coefficient of x3x^3 is $(1-2

2)\sqrt{2}). On the right side, the coefficient of x3x^3 is (1−a)(1-a). Equating these, we get our first equation:

1 - 2 $\sqrt{2} = 1 - a

2. Coefficients of x2x^2: On the left side, the coefficient of x2x^2 is −3-3. On the right side, the coefficient of x2x^2 is −3-3. Equating these gives: $ -3 = -3 $ This equation is always true and doesn't give us any new information about aa, bb, or cc. This often happens in polynomial identities; some coefficients match up automatically, which is a good sign we're on the right track!

3. Coefficients of x1x^1 (or just xx): On the left side, the coefficient of xx is ((3a−b)+(1−c)12)((3a - b) + (1-c)^{\frac{1}{2}}). On the right side, the coefficient of xx is 11. Equating these, we get our second equation:

(3a−b)+(1−c)12=1 (3a - b) + (1-c)^{\frac{1}{2}} = 1

4. Constant Terms: On the left side, the constant term is abab. On the right side, the constant term is −232+3-2^{\frac{3}{2}}+3. Let's simplify the right side's constant term. We know that $2^{\frac{3}{2}} = (23){\frac{1}{2}} = 8^{\frac{1}{2}} = \sqrt{8} = \sqrt{4 \times 2} = 2

2\sqrt{2}. So, the constant term on the right is $-2

2+3\sqrt{2} + 3. Equating the constant terms gives us our third equation:

ab = 3 - 2 $\sqrt{2}

Now we have a system of three equations with three unknowns:

  1. $1 - 2

$\sqrt{2} = 1 - a $ 2. $(3a - b) + (1-c)^{\frac{1}{2}} = 1 $ 3. $ab = 3 - 2

$\sqrt{2} $

This is where the real problem-solving kicks in. We'll use these equations to systematically find the values of aa, bb, and cc. The first equation looks like the easiest to solve, so let's start there!

Solving for aa, bb, and cc

We've successfully set up our system of equations by equating the coefficients. Now comes the satisfying part: solving for the actual values of aa, bb, and cc. Let's tackle them one by one, using the equations we derived:

Step 1: Solve for aa using Equation 1 Our first equation is: $ 1 - 2

$\sqrt{2} = 1 - a $ To solve for aa, we can subtract 1 from both sides:

-2 $\sqrt{2} = -a

Now, multiply both sides by -1:

a = 2 $\sqrt{2}

Step 2: Solve for bb using Equation 3 Our third equation involves both aa and bb: $ ab = 3 - 2

$\sqrt{2} $ We now know the value of aa, so we can substitute $a = 2

2\sqrt{2} into this equation:

(2 $\sqrt{2}) b = 3 - 2 $\sqrt{2}

To find bb, divide both sides by $2

2\sqrt{2}:

b = rac{3 - 2 $\sqrt{2}}{2 $\sqrt{2}}

To simplify this expression for bb, let's rationalize the denominator by multiplying the numerator and denominator by 2\sqrt{2}:

b = rac{(3 - 2 $\sqrt{2}) $\sqrt{2}} {(2 $\sqrt{2}) $\sqrt{2}}}

b = rac{3 $\sqrt{2}} - 2 ( $\sqrt{2})^2} {2 ( $\sqrt{2})^2}

b = rac{3 $\sqrt{2}} - 2(2)} {2(2)}

b = rac{3 $\sqrt{2}} - 4} {4}

So, we have $b = rac{3

\sqrt{2}} - 4}{4}. This looks a bit messy, but it's the correct value for bb. Let's keep going!

Step 3: Solve for cc using Equation 2 Our second equation involves aa, bb, and cc: $ (3a - b) + (1-c)^{\frac{1}{2}} = 1 $ We know $a = 2

2\sqrt{2} and $b = rac{3

\sqrt{2}} - 4}{4}. Let's substitute these values into the equation. First, let's calculate 3a−b3a - b:

3a - b = 3(2 $\sqrt{2}) - rac{3 $\sqrt{2}} - 4}{4}

3a - b = 6 $\sqrt{2} - rac{3 $\sqrt{2}} - 4}{4}

To subtract these, we need a common denominator:

3a - b = rac{4(6 $\sqrt{2})} {4} - rac{3 $\sqrt{2}} - 4}{4}

3a - b = rac{24 $\sqrt{2}} - (3 $\sqrt{2}} - 4)} {4}

3a - b = rac{24 $\sqrt{2}} - 3 $\sqrt{2}} + 4} {4}

3a - b = rac{21 $\sqrt{2}} + 4} {4}

Now substitute this back into Equation 2:

rac{21 $\sqrt{2}} + 4} {4} + (1-c)^{\frac{1}{2}} = 1

Isolate the term with cc:

(1-c)^{\frac{1}{2}} = 1 - rac{21 $\sqrt{2}} + 4} {4}

Get a common denominator on the right side:

(1-c)^{\frac{1}{2}} = rac{4}{4} - rac{21 $\sqrt{2}} + 4} {4}

(1-c)^{\frac{1}{2}} = rac{4 - (21 $\sqrt{2}} + 4)} {4}

(1-c)^{\frac{1}{2}} = rac{4 - 21 $\sqrt{2}} - 4} {4}

(1-c)^{\frac{1}{2}} = rac{-21 $\sqrt{2}}}{4}

This is where we hit a snag, guys. The left side, (1−c)12(1-c)^{\frac{1}{2}}, represents a square root, which typically yields a non-negative real number when the expression inside is non-negative. However, the right side, $\frac{-21

\sqrt{2}}}{4}, is a negative number. In the realm of real numbers, the square root of a number cannot be negative. This implies that there might be an issue with the original problem statement or that we need to consider complex numbers. However, usually in these types of problems, we expect real solutions. Let's double-check our calculations.

Revisiting the Expansion and Simplification Let's re-verify the expansion of (a−x)(3x+b)(a-x)(3x+b). (a−x)(3x+b)=3ax+ab−3x2−bx=−3x2+(3a−b)x+ab(a-x)(3x+b) = 3ax + ab - 3x^2 - bx = -3x^2 + (3a-b)x + ab. This is correct. Combining terms on the left side:

(1-2 $\sqrt{2}) x^3 - 3x^2 + ((3a-b) + (1-c)^{\frac{1}{2}})x + ab$. This is also correct. *Revisiting the Right Side* $(1-a) x^3-3 x^2+x-2^{\frac{3}{2}}+3$ $2^{\frac{3}{2}} = 2 $\sqrt{2}$. So the right side is $(1-a) x^3-3 x^2+x+3 - 2 $\sqrt{2}$. This is correct. *Revisiting Coefficient Equations* $x^3: 1 - 2 $\sqrt{2} = 1 - a \implies a = 2 $\sqrt{2}$. Correct. $x^2: -3 = -3$. Correct. $x^1: (3a-b) + (1-c)^{\frac{1}{2}} = 1$. Correct. Constant: $ab = 3 - 2 $\sqrt{2}$. Correct. *Revisiting Calculation for $b$* $a = 2 $\sqrt{2}$. $ab = 3 - 2 $\sqrt{2}$ $(2 $\sqrt{2})b = 3 - 2 $\sqrt{2}$ $b = rac{3 - 2 $\sqrt{2}}{2 $\sqrt{2}} = rac{(3 - 2 $\sqrt{2}) $\sqrt{2}}}{2( $\sqrt{2})^2} = rac{3 $\sqrt{2}} - 4}{4}$. Correct. *Revisiting Calculation for $c$* $3a - b = 3(2 $\sqrt{2}) - rac{3 $\sqrt{2}} - 4}{4} = 6 $\sqrt{2} - rac{3 $\sqrt{2}} - 4}{4} = rac{24 $\sqrt{2}} - (3 $\sqrt{2}} - 4)}{4} = rac{21 $\sqrt{2}} + 4}{4}$. Correct. $(3a - b) + (1-c)^{\frac{1}{2}} = 1$ $ rac{21 $\sqrt{2}} + 4}{4} + (1-c)^{\frac{1}{2}} = 1$ $(1-c)^{\frac{1}{2}} = 1 - rac{21 $\sqrt{2}} + 4}{4} = rac{4 - (21 $\sqrt{2}} + 4)}{4} = rac{-21 $\sqrt{2}}}{4}$. It appears that with the given polynomial identity and assuming real number solutions for $a, b, c$ and the terms involving them, there might not be a valid solution for $c$. The equation $(1-c)^{\frac{1}{2}} = rac{-21 $\sqrt{2}}}{4}$ has no real solution for $c$ because the square root of a real number cannot be negative. If we were working with complex numbers, we could proceed, but typically these problems are set within the real number system unless otherwise specified. It's possible there was a typo in the original problem statement that led to this situation. For instance, if the right side constant term was $2^{\frac{3}{2}}+3$ instead of $-2^{\frac{3}{2}}+3$, or if there was a negative sign in front of the $x$ term on the left, we might get a valid real solution for $c$. However, if we strictly follow the given problem and are allowed complex numbers: Square both sides: $1-c = igg( rac{-21 $\sqrt{2}}}{4}igg)^2$ $1-c = rac{(-21)^2 ( $\sqrt{2})^2}{4^2}$ $1-c = rac{441 *2}{16}$ $1-c = rac{882}{16}$ $1-c = rac{441}{8}$ $c = 1 - rac{441}{8}$ $c = rac{8}{8} - rac{441}{8}$ $c = rac{8 - 441}{8}$ $c = rac{-433}{8}$ So, if complex numbers are permitted for the value of the square root term, the values are: $a = 2 $\sqrt{2}$ $b = rac{3 $\sqrt{2}} - 4}{4}$ $c = rac{-433}{8}$ **Conclusion:** After a thorough algebraic breakdown, we found the values for $a$ and $b$. The value for $c$ presents an interesting case. If we restrict ourselves to real numbers, the problem as stated has no solution for $c$ because it leads to the square root of a real number equaling a negative value. However, if we allow for complex number interpretations of the square root, we can find a value for $c$. The values are $a = 2 $\sqrt{2}$, $b = rac{3 $\sqrt{2}} - 4}{4}$, and $c = rac{-433}{8}$ (under the complex number interpretation). It's always a good idea to check the context of such problems; usually, if complex numbers are expected, it's mentioned. For standard high school or early college algebra, this would indicate no real solution. But hey, math is full of surprises, right? Hope you guys enjoyed this deep dive!