Question

Let f(x) be a polynomial of degree 5 with integral coefficient that has atleast one integral root. If f(2) = 13 and f(10) = 5, then value of '$\alpha$' such that f($\alpha$) = 0 (where $\alpha$ $\in$ I) 7(1) 1-0 are real and positive

Answer 1

15

Answer 2

To solve this problem, we use a fundamental property of polynomials with integral coefficients.

Property: If $P(x)$ is a polynomial with integral coefficients, then for any two distinct integers $a$ and $b$, $(a-b)$ divides $(P(a) - P(b))$.

Let $f(x)$ be the given polynomial of degree 5 with integral coefficients. We are given that $f(x)$ has at least one integral root, let's call it $\alpha$. So, $f(\alpha) = 0$, where $\alpha \in \mathbb{Z}$.

We are also given:

1. $f(2) = 13$
2. $f(10) = 5$

Now, we apply the property mentioned above:

Step 1: Using $f(2)$ and $f(\alpha)$ Since $f(x)$ has integral coefficients, $(2 - \alpha)$ must divide $(f(2) - f(\alpha))$. Substituting the given values: $(2 - \alpha)$ divides $(13 - 0)$ $(2 - \alpha)$ divides $13$.

The integral divisors of 13 are $\pm 1, \pm 13$. So, $2 - \alpha$ can be $1, -1, 13, -13$.

• If $2 - \alpha = 1 \implies \alpha = 2 - 1 = 1$
• If $2 - \alpha = -1 \implies \alpha = 2 - (-1) = 3$
• If $2 - \alpha = 13 \implies \alpha = 2 - 13 = -11$
• If $2 - \alpha = -13 \implies \alpha = 2 - (-13) = 15$ Possible integral values for $\alpha$ from this condition are $\{1, 3, -11, 15\}$.

Step 2: Using $f(10)$ and $f(\alpha)$ Similarly, $(10 - \alpha)$ must divide $(f(10) - f(\alpha))$. Substituting the given values: $(10 - \alpha)$ divides $(5 - 0)$ $(10 - \alpha)$ divides $5$.

The integral divisors of 5 are $\pm 1, \pm 5$. So, $10 - \alpha$ can be $1, -1, 5, -5$.

• If $10 - \alpha = 1 \implies \alpha = 10 - 1 = 9$
• If $10 - \alpha = -1 \implies \alpha = 10 - (-1) = 11$
• If $10 - \alpha = 5 \implies \alpha = 10 - 5 = 5$
• If $10 - \alpha = -5 \implies \alpha = 10 - (-5) = 15$ Possible integral values for $\alpha$ from this condition are $\{9, 11, 5, 15\}$.

Step 3: Find the common value of $\alpha$ For $\alpha$ to be an integral root of $f(x)$, it must satisfy both conditions simultaneously. We look for the common value(s) in the two sets of possible $\alpha$ values: Set 1: $\{1, 3, -11, 15\}$ Set 2: $\{9, 11, 5, 15\}$

The only common value is $\alpha = 15$.

Thus, the value of the integral root $\alpha$ is 15. The additional text "7(1) 1-0 are real and positive" appears to be irrelevant to the polynomial problem and is ignored.

The final answer is $\boxed{15}$.

Explanation of the solution: The problem leverages the property that for a polynomial $f(x)$ with integral coefficients, if $a$ and $b$ are integers, then $(a-b)$ divides $(f(a) - f(b))$. Let $\alpha$ be an integral root, so $f(\alpha) = 0$.

1. Using $f(2) = 13$, we have $(2 - \alpha)$ divides $(f(2) - f(\alpha))$, which means $(2 - \alpha)$ divides $(13 - 0) = 13$. This implies $(2 - \alpha) \in \{\pm 1, \pm 13\}$, leading to $\alpha \in \{1, 3, -11, 15\}$.
2. Using $f(10) = 5$, we have $(10 - \alpha)$ divides $(f(10) - f(\alpha))$, which means $(10 - \alpha)$ divides $(5 - 0) = 5$. This implies $(10 - \alpha) \in \{\pm 1, \pm 5\}$, leading to $\alpha \in \{9, 11, 5, 15\}$. The only value common to both sets of possible $\alpha$ values is 15.