Problem 12
Question
A polynomial \(P\) is given. (a) Find all zeros of \(P\), real and complex. (b) Factor \(P\) completely. $$P(x)=x^{4}+6 x^{2}+9$$
Step-by-Step Solution
Verified Answer
The zeros are \( x = i\sqrt{3}, -i\sqrt{3} \) and the factorization is \( P(x) = (x^2 + 3)^2 \).
1Step 1: Recognize the form of the polynomial
The given polynomial is \(P(x) = x^4 + 6x^2 + 9\). Notice that the polynomial is a quadratic in form if we set \( y = x^2 \), transforming it to \( y^2 + 6y + 9 \).
2Step 2: Substitute and solve the quadratic
Substitute \( y = x^2 \) into the polynomial, yielding the equation \( y^2 + 6y + 9 = 0 \). This can be factored as \( (y + 3)(y + 3) = 0 \) or \( (y + 3)^2 = 0 \).
3Step 3: Find the roots of the quadratic
From the factored form \( (y + 3)^2 = 0 \), we find the solution \( y = -3 \). Thus, \( x^2 = -3 \).
4Step 4: Solve for x
If \( x^2 = -3 \), then \( x = \pm i\sqrt{3} \) where \(i\) is the imaginary unit. These are the zeros of the polynomial: \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \).
5Step 5: Factor the polynomial completely
Use the zeros to write the polynomial as a product of linear factors in terms of \( x \): \( P(x) = (x - i\sqrt{3})^2 (x + i\sqrt{3})^2 \). This uses the fact that \( (x^2 + 3) = (x - i\sqrt{3})(x + i\sqrt{3}) \). Therefore, \( P(x) = (x^2 + 3)^2 \).
Key Concepts
Zeros of a PolynomialFactoring PolynomialsComplex Numbers
Zeros of a Polynomial
Finding the zeros of a polynomial involves determining the values of the variable, typically represented as \( x \), that make the polynomial equal to zero. These values are also known as the roots of the polynomial.
For our polynomial \( P(x) = x^4 + 6x^2 + 9 \), we first need to express it in a form that allows us to solve it. By setting \( y = x^2 \), the polynomial becomes a quadratic: \( y^2 + 6y + 9 \).
Solving the quadratic equation \( y^2 + 6y + 9 = 0 \) by factoring, we get \( (y + 3)^2 = 0 \), which implies \( y = -3 \). Since \( y = x^2 \), we solve \( x^2 = -3 \), leading to zeros: \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \).
This involves complex numbers because the square root of a negative number leads to the imaginary unit \( i \).
For our polynomial \( P(x) = x^4 + 6x^2 + 9 \), we first need to express it in a form that allows us to solve it. By setting \( y = x^2 \), the polynomial becomes a quadratic: \( y^2 + 6y + 9 \).
Solving the quadratic equation \( y^2 + 6y + 9 = 0 \) by factoring, we get \( (y + 3)^2 = 0 \), which implies \( y = -3 \). Since \( y = x^2 \), we solve \( x^2 = -3 \), leading to zeros: \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \).
This involves complex numbers because the square root of a negative number leads to the imaginary unit \( i \).
Factoring Polynomials
Factoring a polynomial means breaking it down into simpler polynomials (of lower degrees), whose product results in the original polynomial. This is crucial as it plays a big role in solving polynomial equations and finding polynomial roots.
For \( P(x) = x^4 + 6x^2 + 9 \), once we determined the polynomial is a disguised quadratic in terms of \( y = x^2 \), it becomes straightforward to factor. We determined that \( (y + 3)^2 = 0 \) gives us \( y + 3 = 0 \) or \( y = -3 \).
Substituting back, \( x^2 = -3 \) leads us to factor the polynomial as \( (x^2 + 3)^2 \). Essentially, factoring involves realizing \((x^2 + 3) = (x - i\sqrt{3})(x + i\sqrt{3})\), creating the factorization \( P(x) = (x - i\sqrt{3})^2 (x + i\sqrt{3})^2 \). This approach utilizes complex numbers implicitly because of the need to incorporate \( i \) when solving for \( x \).
For \( P(x) = x^4 + 6x^2 + 9 \), once we determined the polynomial is a disguised quadratic in terms of \( y = x^2 \), it becomes straightforward to factor. We determined that \( (y + 3)^2 = 0 \) gives us \( y + 3 = 0 \) or \( y = -3 \).
Substituting back, \( x^2 = -3 \) leads us to factor the polynomial as \( (x^2 + 3)^2 \). Essentially, factoring involves realizing \((x^2 + 3) = (x - i\sqrt{3})(x + i\sqrt{3})\), creating the factorization \( P(x) = (x - i\sqrt{3})^2 (x + i\sqrt{3})^2 \). This approach utilizes complex numbers implicitly because of the need to incorporate \( i \) when solving for \( x \).
Complex Numbers
Complex numbers arise naturally when dealing with polynomial equations, especially when solutions extend beyond the real number domain. A complex number combines a real part and an imaginary part, and is typically written as \( a + bi \), where \( i \) is the imaginary unit, \( i^2 = -1 \).
When solving \( x^2 = -3 \), the result translates into complex solutions: \( x = \pm i\sqrt{3} \). The solutions are not just ordinary numbers but complex, as \( i \) accounts for the presence of negative under square roots.
Understanding complex numbers is critical for solving problems where real numbers do not suffice. In polynomial equations like \( P(x) = x^4 + 6x^2 + 9 \), complex roots ensure that all possible solutions are accounted for, preserving the fundamental theorem of algebra, which states every polynomial equation of degree \( n \) must have exactly \( n \) roots, considering multiplicity. In summary, complex numbers extend our computational horizon beyond the real number line.
When solving \( x^2 = -3 \), the result translates into complex solutions: \( x = \pm i\sqrt{3} \). The solutions are not just ordinary numbers but complex, as \( i \) accounts for the presence of negative under square roots.
Understanding complex numbers is critical for solving problems where real numbers do not suffice. In polynomial equations like \( P(x) = x^4 + 6x^2 + 9 \), complex roots ensure that all possible solutions are accounted for, preserving the fundamental theorem of algebra, which states every polynomial equation of degree \( n \) must have exactly \( n \) roots, considering multiplicity. In summary, complex numbers extend our computational horizon beyond the real number line.
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