82. There is exactly one polynomial with integer coefficients with zeros of 2, 4, and 6.

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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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**Problem 82:**

There is exactly one polynomial with integer coefficients with zeros of 2, 4, and 6.

**Explanation:**

To find a polynomial with given zeros, we can use the fact that if a polynomial has zeros at \( x = a \), \( x = b \), and \( x = c \), then the polynomial can be expressed in the factored form:

\[
P(x) = k(x - a)(x - b)(x - c)
\]

where \( k \) is a nonzero constant.

For the zeros 2, 4, and 6, and since we want integer coefficients, we can set \( k = 1 \). Therefore, the polynomial is:

\[
P(x) = (x - 2)(x - 4)(x - 6)
\]

Expanding this, we get the polynomial:

\[
P(x) = (x - 2)(x^2 - 10x + 24) = x^3 - 10x^2 + 24x - 2x^2 + 20x - 48
\]

Simplifying, the polynomial with integer coefficients is:

\[
P(x) = x^3 - 12x^2 + 44x - 48
\]

Thus, the polynomial \( P(x) = x^3 - 12x^2 + 44x - 48 \) is the polynomial with integer coefficients that has zeros at 2, 4, and 6.
Transcribed Image Text:**Problem 82:** There is exactly one polynomial with integer coefficients with zeros of 2, 4, and 6. **Explanation:** To find a polynomial with given zeros, we can use the fact that if a polynomial has zeros at \( x = a \), \( x = b \), and \( x = c \), then the polynomial can be expressed in the factored form: \[ P(x) = k(x - a)(x - b)(x - c) \] where \( k \) is a nonzero constant. For the zeros 2, 4, and 6, and since we want integer coefficients, we can set \( k = 1 \). Therefore, the polynomial is: \[ P(x) = (x - 2)(x - 4)(x - 6) \] Expanding this, we get the polynomial: \[ P(x) = (x - 2)(x^2 - 10x + 24) = x^3 - 10x^2 + 24x - 2x^2 + 20x - 48 \] Simplifying, the polynomial with integer coefficients is: \[ P(x) = x^3 - 12x^2 + 44x - 48 \] Thus, the polynomial \( P(x) = x^3 - 12x^2 + 44x - 48 \) is the polynomial with integer coefficients that has zeros at 2, 4, and 6.
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