Problem 70 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find a polynomial function that has the given zeros. (There are many correct answers.) \(-2,-1,0,1,2\)

Step-by-Step Solution

Answer

One possible polynomial function with the given roots is \(f(x) = (x + 2)(x + 1)(x)(x - 1)(x - 2)\).

Step 1: Understand root to polynomial relationship

Understand that a polynomial can be written in terms of its roots. If \(r_1, r_2, r_3, ...\), are the roots of the polynomial function, then it can be written as \(f(x) = a(x - r_1)(x - r_2)(x - r_3)...\). Here, 'a' can be any real number. To keep things simple, we will use a = 1.

Step 2: Incorporate given roots into the formula

We were given the roots as -2, -1, 0, 1, 2. Following the formula obtained from step 1, we replace the roots in the polynomial function formula. Hence, the function becomes \(f(x) = (x - (-2))(x - (-1))(x - 0)(x - 1)(x - 2)\).

Step 3: Simplify the polynomial

Simplify the expression obtained in step 2. It simplifies to \(f(x) = (x + 2)(x + 1)(x)(x - 1)(x - 2)\).

Key Concepts

Roots of Polynomials

The roots of a polynomial are the values of the variable that make the polynomial equal to zero. For example, if you're given the zeros

• -2
• -1
• 0
• 1
• 2

this means plugging these values into the polynomial yields zero.
To construct a polynomial from these roots, you can start by creating factors out of each root. For each root \(r\), there's a corresponding factor \((x - r)\).
Thus with the given roots, the factors would be:

• \((x - (-2)) = (x + 2)\)
• \((x - (-1)) = (x + 1)\)
• \((x - 0) = x\)
• \((x - 1)\)
• \((x - 2)\)

Putting it all together, the polynomial can be expressed as \(f(x) = (x + 2)(x + 1)x(x - 1)(x - 2)\).
These factors form the skeleton of the polynomial and let you easily see the roots again.

Simplifying Polynomials

Once the polynomial is represented in its unfactored form by using the roots, simplification often helps whether for ease of calculation or pattern recognition. The process involves basic algebra; here’s generally how it works.
Start by multiplying two factors at a time. For instance, given the expression \((x + 2)(x + 1)(x)(x - 1)(x - 2)\), you could start by simplifying a pair. Simplifying \((x + 2)(x + 1)\) gives\[x^2 + 3x + 2\]. Next, you might simplify \((x - 1)(x - 2)\) to \[x^2 - 3x + 2\], and then multiply the results by each other, keeping the last \(x\) aside for the final step.
Eventually, you apply the distributive property

• Multiply terms without missing one
• Combine like terms when applicable

After completing these steps, you eventually obtain a simplified polynomial form.
Before wrapping up, you verify your result by ensuring plugging the roots back into your expression still yields zero.

Factorization of Polynomials

Factorization is the technique of breaking down a polynomial into simpler factors that, when multiplied, give the original polynomial. Initially, you might encounter a lengthy polynomial that is difficult to manage.
By expressing \(x^5 - 10x^3 + 9x\), observe that \(x\) is involved in every term, and therefore can be factored out as a common factor | producing \(x(x^4 - 10x^2 + 9)\).
Beyond factoring out common terms like \(x\), you could further discover quadratic or pattern-based factors.

• If a polynomial can split into quadratic factors, then set it as a quadratic \(\text{(i.e., )}(ax^2 + bx + c)\).
• For complex factorizations, it can require recognition of patterns, like differences of squares or applying the quadratic formula.

Finally finding the simplest components from all factors is highly beneficial. Knowing the roots aids tremendously; it's a guide in constructing and deconstructing your polynomial.