Problem 27 - Algebra and Trigonometry

Question

Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$ P(x)=x^{4}-3 x^{3}+2 x^{2} $$

Step-by-Step Solution

Answer

Zeros are \(x = 0, 1, 2\); graph touches at \(x=0\) and crosses at \(x=1\) and \(x=2\).

Step 1: Factor out the Greatest Common Factor (GCF)

First, identify the greatest common factor in the polynomial. Here, each term has a common factor of \(x^2\). Factoring \(x^2\) out of the polynomial gives:\[P(x) = x^2(x^2 - 3x + 2)\]

Step 2: Factor the Quadratic Expression

Now, focus on factoring the quadratic expression \(x^2 - 3x + 2\). We need to find two numbers that multiply to 2 and add to -3. These numbers are -1 and -2, so we can factor the quadratic as:\[x^2 - 3x + 2 = (x-1)(x-2)\]

Step 3: Write the Fully Factored Form

Using the results from Steps 1 and 2, write the fully factored form of the polynomial:\[P(x) = x^2(x-1)(x-2)\]

Step 4: Find the Zeros of the Polynomial

To find the zeros, set each factor equal to zero and solve:- \(x^2 = 0\) gives \(x = 0\).- \(x-1 = 0\) gives \(x = 1\).- \(x-2 = 0\) gives \(x = 2\).Thus, the zeros of \(P(x)\) are \(x = 0, 1, 2\).

Step 5: Sketch the Graph

The graph of \(P(x) = x^2(x-1)(x-2)\) is a degree 4 polynomial. It will have intercepts at \(x = 0\), \(x = 1\), and \(x = 2\). As \(x o \pm \infty\), the graph will go to positive infinity since the leading coefficient (of \(x^4\)) is positive. At \(x = 0\), the curve touches the x-axis and turns around because it's a double root. At \(x = 1\) and \(x = 2\), the graph crosses the x-axis. Sketch the general shape that has these properties.

Key Concepts

Greatest Common Factor

When you start factoring any polynomial, the first step is to identify the Greatest Common Factor (GCF). This is the largest factor that every term in the polynomial shares. Finding it simplifies the polynomial and makes the next steps more manageable.
Let's look at the given polynomial:

• Terms are: \(x^4\), \(-3x^3\), and \(2x^2\).


• Each term has at least an \(x^2\) in it, making \(x^2\) the GCF.

Once you find the GCF, you factor it out by dividing each term by \(x^2\), resulting in:

• \[P(x) = x^2(x^2 - 3x + 2)\]

With this, you have simplified the polynomial and set up the next steps for further factoring.

Zeros of a Polynomial

Finding the zeros of a polynomial is critical, as these are the values of \(x\) that make the polynomial equal to zero. When you have the polynomial factored, setting each factor equal to zero helps identify these zeros.
In our example, the factored form is:

• \[P(x) = x^2(x-1)(x-2)\]

To find the zeros:

• For \(x^2 = 0\), solving gives \(x = 0\). This is a double root, where the graph will "touch and turn" at zero.


• For \(x-1 = 0\), solving gives \(x = 1\).


• For \(x-2 = 0\), solving gives \(x = 2\).

These are the points where the polynomial crosses or touches the x-axis, giving insights into the shape of its graph.

Quadratic Expression

Quadratic expressions are polynomials of the form \(ax^2 + bx + c\), which can be factored further into simpler binomials. This step is crucial when the quadratic needs factoring for finding zeros.
In the expression we have:

• \(x^2 - 3x + 2\)

The task is to find two numbers that multiply to 2 and add to -3. These numbers are -1 and -2.
This means the quadratic can be factored as:

• \((x-1)(x-2)\)

This factored form is straightforward because it breaks down the expression into more manageable parts, making it easy to identify zeros and helping in graphing the polynomial. Quadratics often play a key role in polynomial problems, so understanding their factoring is essential.