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A polynomial is degree 3, has a root of multiplicity 2 at [tex]x=2[/tex], and a root of multiplicity 1 at [tex]x=-4[/tex]. The [tex]y[/tex]-intercept is [tex]y=-1.6[/tex].

Find a formula for [tex]P(x)[/tex].

[tex]P(x) = \square[/tex]


Sagot :

To find the formula for the polynomial [tex]\( P(x) \)[/tex], we need to follow a methodical approach. We are given several pieces of information about the polynomial, including its roots, multiplicities, and the y-intercept. Here's the detailed step-by-step solution:

1. Identify the roots and their multiplicities:
- The root at [tex]\( x = 2 \)[/tex] has a multiplicity of 2.
- The root at [tex]\( x = -4 \)[/tex] has a multiplicity of 1.

2. Construct the polynomial with these roots and multiplicities:
Given the roots and their multiplicities, the polynomial [tex]\( P(x) \)[/tex] can be expressed in factored form as:
[tex]\[ P(x) = a \cdot (x - 2)^2 \cdot (x + 4) \][/tex]
where [tex]\( a \)[/tex] is a constant coefficient.

3. Use the y-intercept to find the coefficient [tex]\( a \)[/tex]:
The y-intercept is the value of the polynomial when [tex]\( x = 0 \)[/tex]. Given that [tex]\( P(0) = -1.6 \)[/tex], substitute [tex]\( x = 0 \)[/tex] into the polynomial:
[tex]\[ P(0) = a \cdot (0 - 2)^2 \cdot (0 + 4) \][/tex]
Simplify the expression:
[tex]\[ P(0) = a \cdot (4) \cdot 4 \implies P(0) = 16a \][/tex]
We know [tex]\( P(0) = -1.6 \)[/tex], so we set up the equation:
[tex]\[ 16a = -1.6 \][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{-1.6}{16} = -0.1 \][/tex]

4. Substitute [tex]\( a \)[/tex] back into the polynomial:
Now that we have determined [tex]\( a = -0.1 \)[/tex], substitute it back into the polynomial expression:
[tex]\[ P(x) = -0.1 \cdot (x - 2)^2 \cdot (x + 4) \][/tex]

Thus, the formula for [tex]\( P(x) \)[/tex] is:
[tex]\[ P(x) = -0.1 \cdot (x - 2)^2 \cdot (x + 4) \][/tex]