Problem 11 - Algebra Form and Function

Question

The polynomial \(p(x)\) can be written in two forms: I. \(\quad p(x)=2 x^{3}-3 x^{2}-11 x+6\) II. \(p(x)=(x-3)(x+2)(2 x-1)\) Which form most readily shows (a) The zeros of \(p(x) ?\) What are they? (b) The vertical intercept? What is it? (c) The sign of \(p(x)\) as \(x\) gets large, either positive or negative? What are the signs? (d) The number of times \(p(x)\) changes sign as \(x\) increases from large negative to large positive \(x ?\) How many times is this?

Step-by-Step Solution

Answer

Question: Based on the step-by-step solution, which form of the polynomial \(p(x)\) most readily shows: (a) The zeros of the polynomial (b) The vertical intercept (c) The sign as \(x\) gets large, either positive or negative (d) The number of times the polynomial changes its sign Answer: (a) Form II (b) Form I (c) Form I (d) Form II

Step 1: (i) Identify the zeros in Form I

To identify zeros in the first form, we'd need to set the polynomial equal to zero and solve for x: \(2x^3 - 3x^2 - 11x + 6 = 0\). To do so, we need to factorize or use algebraic methods, such as synthetic division, which can be computationally intensive.

Step 2: (ii) Identify the zeros in Form II

In the second form, \(p(x)= (x-3)(x+2)(2x-1)\), we can quickly identify the zeros by setting each factor equal to zero: \(x-3=0\), \(x+2=0\), and \(2x-1=0\).

Step 3: (iii) Most readily shows

Form II most readily shows the zeros of the polynomial as it's easier to solve for the zeros by setting each linear factor equal to zero.

Step 4: (iv) The zeros

Solving the equations, we get \(x = 3\), \(x = -2\), and \(x = \dfrac{1}{2}\). Therefore, the zeros of the polynomial are \(3\), \(-2\), and \(\dfrac{1}{2}\). #b) Vertical intercept #

Step 5: (i) Vertical intercept in Form I

To find the vertical intercept in the first form, all we need to do is plug in \(x=0\): \(p(0)=2(0)^3-3(0)^2-11(0)+6=6\).

Step 6: (ii) Vertical intercept in Form II

In the second form, plug in \(x=0\): \(p(0)=(-3)(2)(-1)=6\).

Step 7: (iii) Most readily shows

Both forms give the vertical intercept without much difficulty. However, Form I is slightly easier since there is no need to multiply terms together.

Step 8: (iv) The vertical intercept

The vertical intercept is \((0, 6)\). #c) Sign of p(x)#

Step 9: (i) Sign in Form I

In the first form, we can analyze the sign of the highest-degree term (\(2x^3\)) for large positive or negative \(x\). As it has an odd exponent, it changes the sign of the polynomial as x gets large, either positive or negative.

Step 10: (ii) Sign in Form II

In the second form, the exponents on each factor are also important. Here, the exponent is just one for each factor, so we examine the product of all three factors.

Step 11: (iii) Most readily shows

Form I most readily shows the sign of \(p(x)\) as \(x\) gets large, either positive or negative because the leading term ultimately determines the sign.

Step 12: (iv) The sign of p(x)

For large positive \(x\), the polynomial has a positive sign. For large negative \(x\), it has a negative sign. #d) Changes in the sign of p(x)#

Step 13: (i) Changes in sign in Form I

In the first form, it's hard to analyze the number of times the polynomial changes sign as \(x\) increases from large negative to large positive values.

Step 14: (ii) Changes in sign in Form II

In the second form, as each factor is linear, we can more easily deduce the changes in the sign of p(x) as x increases.

Step 15: (iii) Most readily shows

Form II most readily shows the number of times the polynomial changes sign as \(x\) increases from large negative to large positive values.

Step 16: (iv) Number of times the sign changes

As x increases, there are two intervals where the polynomial changes its sign: from -2 to 3 and from \(\dfrac{1}{2}\) to 3. Therefore, the polynomial changes its sign two times as x increases from large negative to large positive values.

Key Concepts

Zeros of a Polynomial

To find the zeros of a polynomial, you should look for the values of \(x\) where the polynomial equals zero. A polynomial can often be expressed in a factored form, which makes identifying zeros straightforward. In such cases, each factor of the polynomial, when set to zero, gives a zero of the polynomial. For example, given the expression \(p(x)=(x-3)(x+2)(2x-1)\), setting each factor equal to zero provides solutions for \(x\). Here, the solutions, or zeros, are:

• \(x - 3 = 0 \Rightarrow x = 3 \)
• \(x + 2 = 0 \Rightarrow x = -2 \)
• \(2x - 1 = 0 \Rightarrow x = \dfrac{1}{2} \)

Thus, the zeros of the polynomial are \(x = 3\), \(x = -2\), and \(x = \dfrac{1}{2}\). These zeros are easily found in the factored form of the polynomial.

Vertical Intercept

The vertical intercept of a polynomial is the value where the graph of the polynomial crosses the y-axis. To find this intercept, evaluate the polynomial at \(x = 0\). This means you substitute 0 for \(x\) in the polynomial expression. In the given polynomial forms:

• Form I: \(p(x) = 2x^3 - 3x^2 - 11x + 6\)
• Form II: \(p(x)=(x-3)(x+2)(2x-1)\)

Substitute \(x = 0\) into either form: - From Form I: \(p(0) = 2(0)^3 - 3(0)^2 - 11(0) + 6 = 6\) - From Form II: \(p(0) = (-3)(2)(-1) = 6\) Therefore, the vertical intercept of the polynomial is \((0, 6)\). It's the same in both forms, making it consistent and easy to find.

Sign of Polynomial Function

Understanding the sign of a polynomial as \(x\) becomes very large positively or negatively involves analyzing the leading term of its standard form. The sign is determined primarily by the highest-degree term, since it has the most influence as \(x\) grows large. For the polynomial \(p(x) = 2x^3 - 3x^2 - 11x + 6\), the leading term is \(2x^3\). This term is positive when x is positive, leading to a positive value of the polynomial. Similarly, when x is negative, because the exponent is odd, the polynomial value turns negative. Thus, for large positive \(x\), the polynomial approaches positive infinity, while for large negative \(x\), it goes towards negative infinity.

Factored Form of Polynomials

The factored form of a polynomial is particularly useful in visualizing and computing certain properties of the polynomial, especially its zeros and changes in sign. When a polynomial is expressed as a product of its factors, each factor corresponds to a zero, simplifying the process of solving for roots. The polynomial given in the exercise, \(p(x) = (x-3)(x+2)(2x-1)\), shows its zeros explicitly. This form also makes it easier to analyze changes in sign. As \(x\) crosses each zero, the sign of the polynomial may change.

• As \(x\) moves from negative to positive through \(-2\), \(1/2\), and \(3\), the polynomial's value transitions through intervals where it changes sign.
• The simplicity of the factored form allows analysts to distinctly see where and how the polynomial's sign shifts across these intervals.

These sign changes are critical for determining the behavior of the polynomial across various ranges of \(x\).