Question

the polynomial of degree 3, (p(x)), has a root of multiplicity 2 at (x = 2) and a root of multiplicity 1 at (x = -2). the (y)-intercept is (y = -6.4).
find a formula for (p(x)).
(p(x) = )

Explanation:

🆕 New Concept Discovered: Writing Polynomial Formulas from Roots
Building equations using roots, multiplicities, and a given point.

Step 1: Set up the general form using roots and multiplicities

A polynomial \( P(x) \) with known roots can be written in factored form:

$$ P(x) = a(x - r_1)^{m_1}(x - r_2)^{m_2}\dots $$

where \( a \) is a constant stretch factor, \( r_i \) are the roots, and \( m_i \) are their respective multiplicities.

We are given:

• A root of multiplicity 2 at \( x = 2 \), which gives the factor \( (x - 2)^2 \).
• A root of multiplicity 1 at \( x = -2 \), which gives the factor \( (x - (-2))^1 = (x + 2) \).

Combining these, the general formula for our degree 3 polynomial is:

$$ P(x) = a(x - 2)^2(x + 2) $$

Step 2: Solve for the leading coefficient \( a \)

We are given that the \( y \)-intercept is \( y = -6.4 \). This means that when \( x = 0 \), \( P(0) = -6.4 \).

Substitute \( x = 0 \) into our formula:

$$ -6.4 = a(0 - 2)^2(0 + 2) $$

$$ -6.4 = a(-2)^2(2) $$

$$ -6.4 = a(4)(2) $$

$$ -6.4 = 8a $$

Now, solve for \( a \):

$$ a = \frac{-6.4}{8} $$

$$ a = -0.8 $$

Step 3: Write the final formula

Substitute \( a = -0.8 \) back into the general equation:

$$ P(x) = -0.8(x - 2)^2(x + 2) $$

Answer:

\( P(x) = -0.8(x - 2)^2(x + 2) \)