%5E3-x(3x%2B1)%5E2%2B(2x%2B1)(4x%5E2-2x%2B1)%3D28.jpeg "(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28")
Solving the Equation: (x+3)^3 - x(3x+1)^2 + (2x+1)(4x^2-2x+1) = 28
This article will guide you through solving the equation (x+3)^3 - x(3x+1)^2 + (2x+1)(4x^2-2x+1) = 28.
Expanding the Equation
First, we need to expand the equation to get rid of the parentheses. Let's break it down step-by-step:
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(x+3)^3: This is a cube of a binomial, which can be expanded using the formula (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. Applying this to our case, we get: (x+3)^3 = x^3 + 9x^2 + 27x + 27
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x(3x+1)^2: Here, we have a product of a monomial and a square of a binomial. Expanding the square first: (3x+1)^2 = 9x^2 + 6x + 1 Then multiplying by x: x(3x+1)^2 = 9x^3 + 6x^2 + x
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(2x+1)(4x^2-2x+1): This is a multiplication of two binomials. We can use the FOIL (First, Outer, Inner, Last) method to expand it: (2x+1)(4x^2-2x+1) = 8x^3 - 4x^2 + 2x + 4x^2 - 2x + 1 = 8x^3 + 1
Now, let's substitute these expanded expressions back into the original equation:
x^3 + 9x^2 + 27x + 27 - (9x^3 + 6x^2 + x) + 8x^3 + 1 = 28
Simplifying the Equation
Next, we'll simplify the equation by combining like terms:
x^3 + 9x^2 + 27x + 27 - 9x^3 - 6x^2 - x + 8x^3 + 1 = 28 (-9x^3 + x^3 + 8x^3) + (9x^2 - 6x^2) + (27x - x) + (27 + 1) = 28 0x^3 + 3x^2 + 26x + 28 = 28
Solving for x
Finally, we have a quadratic equation. To solve it, we can move all terms to one side:
3x^2 + 26x = 0
Now, factor out the common factor of x:
x(3x + 26) = 0
This gives us two possible solutions:
- x = 0
- 3x + 26 = 0 => x = -26/3
Conclusion
Therefore, the solutions to the equation (x+3)^3 - x(3x+1)^2 + (2x+1)(4x^2-2x+1) = 28 are x = 0 and x = -26/3.