Problem 11 - Precalculus : Building Concepts and Connections

Question

Find all the zeros, real and nonreal, of the polynomial. Then express \(p(x)\) as a product of linear factors. $$p(x)=2 x^{2}-5 x+3$$

Step-by-Step Solution

Answer

Applying the quadratic formula gives the roots of the equation as \(\frac{3}{2}\) and \(\frac{1}{2}\). Therefore, the polynomial can be expressed as a product of linear factors as \(2*(x - \frac{3}{2})*(x - \frac{1}{2})\).

Step 1: Apply the Quadratic Formula

Applying the quadratic formula, which is \( x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}} \), to the given equation \(2x^2 - 5x + 3 = 0\), where a = 2, b = -5, and c = 3, the roots of the equation can be obtained.

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). Calculate this for our specific equation.

Step 3: Determine the Roots

Using the discriminant value, calculate the roots of the equation. It should be noted that if the discriminant is less than zero, the roots are nonreal (complex numbers). If the discriminant is equal to zero, there is one real root (repeat root), and if the discriminant is greater than zero, there are two distinct real roots.

Step 4: Express the Polynomial as a Product of Linear Factors

After obtaining the roots of the polynomial, express it as a product of linear factors. A quadratic polynomial will generally factor as \(a*(x - r1)*(x - r2)\) where \(r1\) and \(r2\) are the roots of the polynomial.

Key Concepts

Quadratic Formula

When faced with a quadratic polynomial like \(p(x) = 2x^2 - 5x + 3\), finding the zeros can be done efficiently using the quadratic formula. This powerful tool comes in handy when factoring by inspection is not straightforward. The formula is expressed as follows:

• \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \)

For your polynomial, identify the coefficients: \(a = 2\), \(b = -5\), and \(c = 3\). By substituting these values, you can solve for the zeros, or roots, of the polynomial. Remember, the plus-minus sign in the formula means you will get two solutions, which denote the zeros of the polynomial. This formula is especially useful because it works for any type of quadratic polynomial, regardless of whether the roots are real or complex.
Applying the quadratic formula, you either end up with one or two solutions. These very solutions are crucial for understanding the polynomial's behavior and its graph.

Discriminant

The discriminant is a key component in the quadratic formula which helps us predict the nature of the roots without solving the entire equation. You calculate the discriminant using the expression \(b^2 - 4ac\). It acts like a detective, revealing whether the polynomial has real or nonreal roots.

• If it’s positive, the polynomial has two distinct real roots.
• If zero, the polynomial has one real repeated root.
• If negative, the roots are complex and nonreal numbers.

For \(2x^2 - 5x + 3\), calculate the discriminant:\(( -5)^2 - 4\cdot2\cdot3 \).Calculate to find:\(25 - 24 = 1\).A positive discriminant here tells us it will have two real distinct roots! Not only does this guide us in expecting real solutions but also helps in sketching the graph and understanding the characteristics of the quadratic polynomial.

Linear Factors

Once the roots are known, the next step is to express the polynomial as a product of linear factors. This is essentially reformulating the polynomial into a multiplication of binomials, which is much like breaking it down to its simplest components.For a quadratic, if the roots are \(r_1\) and \(r_2\), the factorization normally takes the form:\( a(x - r_1)(x - r_2) \).Applying this to our distinct roots from the quadratic formula, you substitute them back into this format, ensuring to include the value of \(a\) from the original quadratic equation.In our example, once roots are found and confirmed, if they were for instance, \(x = 3\) and \(x = \frac{1}{2}\), the expression changes to:\(2(x - 3)(x - \frac{1}{2})\).Expressing it in this way not only verifies the polynomial’s roots but also is a compact representation of the original expression, making it easy to understand and further analyze the polynomial's behavior.