Problem 30 - Precalculus with Calculus Previews

Question

Find all real zeros of the given polynomial function \(f\). Then factor \(f(x)\) using only real numbers. $$ f(x)=x^{6}-12 x^{4}+48 x^{2}-64 $$

Step-by-Step Solution

Answer

The real zeros of \( f(x) \) are \( x = 2 \) and \( x = -2 \). The factored form is \( (x - 2)^3(x + 2)^3 \).

Step 1: Simplify the Polynomial

First, observe the structure of the polynomial. Notice that each term has a power of 2. Thus, substitute \( y = x^2 \). Then the polynomial becomes \( f(y) = y^3 - 12y^2 + 48y - 64 \).

Step 2: Identify Possible Rational Roots

Apply the Rational Root Theorem to \( f(y) \). The possible rational roots are the factors of the constant term \(-64\), which include \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm 64\).

Step 3: Check for Real Roots

By testing these possible roots, we find that \( y = 4 \) is a root of \( f(y) \). Use long division or synthetic division to divide \( f(y) \) by \( y - 4 \).

Step 4: Factor the Cubic Polynomial

Dividing \( f(y) \) by \( y - 4 \) results in \( (y - 4)(y^2 - 8y + 16) \). Notice that \( y^2 - 8y + 16 \) can be rewritten as \( (y - 4)^2 \). Thus, \( f(y) \) factors into \( (y - 4)^3 \).

Step 5: Convert Back to Original Variable

Since \( y = x^2 \), substitute back to get \( f(x) = (x^2 - 4)^3 \). Recognize \( x^2 - 4 \) is a difference of squares: \( (x - 2)(x + 2) \).

Step 6: Find All Real Zeros of \( f(x) \)

The zeros of \( x^2 - 4 = 0 \) are \( x = 2 \) and \( x = -2 \). These are the zeros of \( f(x) \).

Step 7: Conclude with Polynomial Factoring

Thus, the factored form of \( f(x) \) is \( (x - 2)^3(x + 2)^3 \).

Key Concepts

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial expression into simpler, lower-degree polynomials that can be multiplied together to obtain the original polynomial.
When we factor a polynomial, we aim to express it as a product of its factors. Each factor is typically a simpler polynomial.

• Factoring helps in solving polynomial equations: By expressing a polynomial as the product of its factors, we make it easier to find its roots.
• For example, if we have a quadratic polynomial like \(x^2 - 4\), we can factor it into \((x-2)(x+2)\).
• In our exercise, we began with the polynomial \(f(x) = x^6 - 12x^4 + 48x^2 - 64\). We factored it step-by-step, eventually resulting in \((x-2)^3(x+2)^3\).

Factoring can sometimes involve recognizing patterns such as the difference of squares or perfect square trinomials, making complex processes more approachable.

Rational Root Theorem

The Rational Root Theorem is a useful tool that helps us identify possible rational roots of a polynomial equation.
It states that for a polynomial equation with integer coefficients, any rational solution or root \(\frac{p}{q}\) is such that \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient.

• In our example, we look at the polynomial \(f(y) = y^3 - 12y^2 + 48y - 64\) after substituting \(y = x^2\).
• The constant term is -64, so all factors (\(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm 64\)) are potential rational roots.
• Checking these possibilities helps us identify actual roots, such as \(y = 4\).

Understanding this theorem enables us to narrow down the list of potential roots to test, making the process of solving polynomials more efficient.

Synthetic Division

Synthetic division is a simplified form of polynomial long division, primarily used for dividing polynomials by linear divisors of the form \((x - c)\).
This method is particularly efficient because it reduces the complexity of the division process without losing accuracy.

• In synthetic division, we focus on coefficients only, allowing us to quickly evaluate whether a number is a root of a polynomial.
• For our function \(f(y)\), after positing \(y=4\) as a root, synthetic division confirmed it by zeroing the remainder, thus dividing \(f(y)\) without performing long division.
• Upon dividing \(f(y)\) by \(y-4\), we simplified it to \((y-4)^3\).

Synthetic division is not only a time-saver but also a clear tool for verifying roots, crucial in polynomial factoring.

Difference of Squares

The difference of squares is a specific type of factoring pattern that applies when a polynomial can be expressed as \(a^2 - b^2\).
This expression can be factored as \((a+b)(a-b)\), simplifying many quadratic forms.

• In our polynomial, \(x^2 - 4\) is written as a difference of squares: \((x-2)(x+2)\).
• This pattern helps us easily find real zeros. For \(x^2 - 4 = 0\), the zeros are \(x = 2\) and \(x = -2\).
• Recognizing this pattern saves effort and provides a clear path to factor and solve equations quickly.

Using the difference of squares fact, we rapidly break down part of our polynomial, emphasizing the power of pattern recognition in mathematics.