Problem 15 - College Algebra with Corequisite Support

Question

For the following exercises, use the Factor Theorem to find all real zeros for the given polynomial function and one factor. $$ f(x)=2 x^{3}+x^{2}-5 x+2 ; \quad x+2 $$

Step-by-Step Solution

Answer

The real zeros are \(-2, 1, \frac{1}{2}\).

Step 1: Recall the Factor Theorem

The Factor Theorem states that if \( (x - c) \) is a factor of a polynomial \( f(x) \), then \( c \) is a zero (or root) of \( f(x) \). Conversely, if \( f(c) = 0 \), then \( (x - c) \) is a factor of \( f(x) \). Here, we know \( x + 2 \) is a factor, so \( x = -2 \) is a root.

Step 2: Synthetic Division

We'll use synthetic division to divide the polynomial \( f(x) = 2x^3 + x^2 - 5x + 2 \) by \( x + 2 \), confirming that \( x = -2 \) is a root and finding other factors. Write the coefficients \([2, 1, -5, 2]\) and perform synthetic division with \(-2\) (the root).

Step 3: Execute Synthetic Division

Perform the division: \( \begin{array}{r|rrrr}-2 & 2 & 1 & -5 & 2 \ & & -4 & 6 & -2 \\hline & 2 & -3 & 1 & 0 \\end{array}\)The remainder is 0, confirming \( x + 2 \) is a factor. The quotient is \( 2x^2 - 3x + 1 \).

Step 4: Factor the Quotient

Now we need to factor the quadratic polynomial \( 2x^2 - 3x + 1 \). We look for two numbers that multiply to \( 2 \cdot 1 = 2 \) and add to \(-3\). These numbers are \(-1\) and \(-2\). We can factor it as \((2x - 1)(x - 1)\).

Step 5: Find All Real Zeros

Set each factor equal to zero to find the roots: 1. \( x + 2 = 0 \) gives \( x = -2 \).2. \( 2x - 1 = 0 \) leads to \( x = \frac{1}{2} \).3. \( x - 1 = 0 \) gives \( x = 1 \). So, the real zeros are \( x = -2, \frac{1}{2}, 1 \).

Key Concepts

Polynomial Zeros

When working with polynomials, one key concept is finding the zeros. Polynomial zeros, also known as roots, are the values of \( x \) for which the polynomial \( f(x) \) equals zero. These are important because they are the points where the graph of the polynomial crosses the x-axis. To find zeros, you can use several methods, such as factoring, using the Quadratic Formula for second degree polynomials, or employing the Factor Theorem for higher-degree polynomials.

The Factor Theorem is an essential tool in this context. It tells us that if \( x - c \) is a factor of a polynomial \( f(x) \), then \( c \) is a zero of the polynomial. Conversely, if \( f(c) = 0 \), then \( x - c \) will divide the polynomial without a remainder. In our exercise, we know \( x + 2 \) is a factor, which means \( x = -2 \) is a zero.

Synthetic Division

Synthetic division is a streamlined method of dividing polynomials when the divisor is of the form \( x - c \). It's simpler and faster than traditional long division, especially for polynomial division. This technique is particularly useful when applying the Factor Theorem to confirm and find zeros.

To use synthetic division, take the following steps:

• Write down the coefficients of the polynomial in a row. For our polynomial \( 2x^3 + x^2 - 5x + 2 \), the coefficients are \( [2, 1, -5, 2] \).
• Place the root \(-2\) outside the division bar – remember this is derived from \( x + 2 \).
• Drop the first coefficient down unchanged.
• Multiply this number by \(-2\), write the result under the next coefficient, and add.
• Continue this multiply and add process across the row.

In the solution, after synthetic division, the quotient is \( 2x^2 - 3x + 1 \) with a remainder of zero, confirming \( x + 2 \) is indeed a factor.

Quadratic Factorization

Quadratic factorization is the process of breaking down a quadratic polynomial into simpler binomials. In quadratic form, a polynomial is expressed as \( ax^2 + bx + c \), and factorization involves finding two binomials \( (px + q) \) and \( (rx + s) \) such that their product equals the original quadratic.

To factor, we seek two numbers that multiply to \( ac \) (the product of the leading coefficient and the constant term) and add to \( b \) (the middle coefficient). For the quadratic \( 2x^2 - 3x + 1 \), these numbers are \(-1\) and \(-2\).

• We can rewrite the middle term \( -3x \) as \( -1x - 2x \).
• Then group and factor by grouping: \( \left(2x^2 - 2x\right) + \left(-1x + 1\right) \).
• Take out the common factors from each group: \( 2x(x - 1) - 1(x - 1) \).
• Notice \( x-1 \) is common, allowing final factorization: \( (2x - 1)(x - 1) \).

With these factors, setting each equal to zero gives us additional roots \( x = \frac{1}{2} \) and \( x = 1 \), completing the list of zeroes for the polynomial.