Solving the Polynomial Equation
We are given the equation: \((x + 1)(x + p)(x^2 + p^2) = x^4 - 1\) Our goal is to find the value of the variable \(p\).
Factoring the Right Side
The term \(x^4 - 1\) on the right side of the equation can be factored using the difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a = x^2\) and \(b = 1\). \(x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)\) Furthermore, the term \(x^2 - 1\) is also a difference of squares, where \(a = x\) and \(b = 1\). So, \(x^2 - 1 = (x - 1)(x + 1)\). Substituting this back, we get: \(x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)\)
Comparing Both Sides
Now, we can rewrite the original equation with the factored right side: \((x + 1)(x + p)(x^2 + p^2) = (x - 1)(x + 1)(x^2 + 1)\) For this equation to hold true for all values of \(x\), the expressions on both sides must be equivalent. If we consider values of \(x\) where \(x \neq -1\), we can cancel the \((x + 1)\) term from both sides: \((x + p)(x^2 + p^2) = (x - 1)(x^2 + 1)\) By comparing the structure of the factors on both sides, we can see a direct match if we set \(p = -1\). Let's substitute \(p = -1\) into the left side expression: \((x + (-1))(x^2 + (-1)^2) = (x - 1)(x^2 + 1)\) This resulting expression exactly matches the right side \((x - 1)(x^2 + 1)\).
Verification
Let's substitute \(p = -1\) back into the original equation to verify: \((x + 1)(x + (-1))(x^2 + (-1)^2) \stackrel{?}{=} x^4 - 1\) \((x + 1)(x - 1)(x^2 + 1) \stackrel{?}{=} x^4 - 1\) Using the difference of squares formula \((a+b)(a-b)=a^2-b^2\): \((x^2 - 1)(x^2 + 1) \stackrel{?}{=} x^4 - 1\) Again, using the difference of squares formula: \((x^2)^2 - 1^2 \stackrel{?}{=} x^4 - 1\) \(x^4 - 1 = x^4 - 1\) The equation holds true. Therefore, the value of \(p\) must be \(-1\).
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