Problem 29 - Precalculus Mathematics for Calculus

Question

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$P(x)=4 x^{4}-25 x^{2}+36$$

Step-by-Step Solution

Answer

Rational zeros: \(\pm 2, \pm \frac{3}{2}\). Factored form: \((x-2)(x+2)(x-\frac{3}{2})(x+\frac{3}{2})\).

Step 1: Identify Possible Rational Zeros

First, identify the possible rational zeros of the polynomial using the Rational Root Theorem. The theorem states that any rational solution of \(P(x)\) has the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term (36) and \(q\) is a factor of the leading coefficient (4). The factors of 36 are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36\), and the factors of 4 are \(\pm 1, \pm 2, \pm 4\). Therefore, the possible rational zeros are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36, \pm \frac{1}{2}, \pm \frac{3}{2}\).

Step 2: Factor the Polynomial as a Quadratic in Disguise

Recognize the polynomial \(P(x) = 4x^4 - 25x^2 + 36\) as a quadratic in terms of \(y = x^2\). Replace \(x^2\) with \(y\) to get \(4y^2 - 25y + 36 = 0\). This is a quadratic equation in \(y\).

Step 3: Solve the Quadratic Equation

Use the quadratic formula to solve for \(y\): \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=4\), \(b=-25\), and \(c=36\). The discriminant is \( (-25)^2 - 4 \times 4 \times 36 = 25\sqrt{49}\). Thus, \(y = \frac{25 \pm 7}{8}\), yielding the solutions \(y = 4\) and \(y = \frac{9}{4}\).

Step 4: Substitute Back in Terms of \(x\)

Since \(y = x^2\), substitute back to solve for \(x\). For \(y = 4\), we have \(x^2 = 4\), so \(x = \pm 2\). For \(y = \frac{9}{4}\), we have \(x^2 = \frac{9}{4}\), so \(x = \pm \frac{3}{2}\).

Step 5: Write the Polynomial in Factored Form

Now, knowing the zeros, we can write \(P(x)\) as a product of its factors: \(P(x) = (x-2)(x+2)(x-\frac{3}{2})(x+\frac{3}{2})\), which is equivalent to the expanded form \(P(x) = (x^2 - 4)(x^2 - \frac{9}{4})\). This can be further expanded to \(P(x) = 4(x^2-4)(x^2-\frac{9}{4})\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool that helps us find all possible rational zeros of a polynomial. If you have a polynomial like \( P(x) = 4x^4 - 25x^2 + 36 \), the theorem tells us that any rational solution \( \frac{p}{q} \) will have \( p \) as a factor of the constant term (here 36) and \( q \) as a factor of the leading coefficient (here 4). This narrows down the list of possibilities significantly.

To apply this:

• Factor 36 to get \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36 \)
• Factor 4 to get \( \pm 1, \pm 2, \pm 4 \)
• Combine these to find all possible rational zeros: \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36, \pm \frac{1}{2}, \pm \frac{3}{2} \)

This theorem doesn't directly provide the solutions but it gives you a smaller selection to check.

Factored Form

Once you find the zeros of a polynomial, you can express it as a product of its factors, which is known as the factored form. This form makes it much easier to see the roots of the polynomial at a glance. For \( P(x) = 4x^4 - 25x^2 + 36 \), by finding the rational roots, we can express it in a more simplified form.

For instance:

• We found the solutions \( x = \pm 2 \) and \( x = \pm \frac{3}{2} \) from the quadratic in disguise method.
• The polynomial can be rewritten in the factored form as \( P(x) = (x-2)(x+2)(x-\frac{3}{2})(x+\frac{3}{2}) \).

This factored form not only offers a compact representation but also directly showcases the solutions.

Quadratic in Disguise

Even though the original polynomial \( P(x) = 4x^4 - 25x^2 + 36 \) looks like a quartic equation, it can be cleverly rewritten to resemble a quadratic. This is why it's called a "quadratic in disguise".

Here's how we recognize this:

• Notice the terms where \( x^4 \) is \( (x^2)^2 \), which suggests substituting \( y = x^2 \).
• This changes the expression to \( 4y^2 - 25y + 36 = 0 \).
• Now, you solve \( 4y^2 - 25y + 36 = 0 \) like an ordinary quadratic equation.

By recognizing this "quadratic in disguise," solving becomes dramatically easier, as typical quadratic methods apply.

Quadratic Formula

The quadratic formula is a key tool for solving quadratic equations. It is expressed as \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our disguised quadratic \( 4y^2 - 25y + 36 = 0 \), we can use it conveniently.

Steps for our problem:

• Identify \( a = 4, b = -25, c = 36 \).
• Calculate the discriminant \( b^2 - 4ac \). It determines the nature of the roots.
• Plug into the formula: \( y = \frac{-(-25) \pm \sqrt{(-25)^2 - 4 \times 4 \times 36}}{2 \times 4} \).
• This yields \( y = 4 \) and \( y = \frac{9}{4} \).

Once you solve for \( y \), reverse the substitution \( y = x^2 \) to find \( x \) values. For \( y = 4 \), we have \( x = \pm 2 \); for \( y = \frac{9}{4} \), we have \( x = \pm \frac{3}{2} \). The quadratic formula provides these concise solutions.