Question

If $(x^{140} + 2x^{151} + K)$ is divisible by $(x+1)$ then the value of K is :

• A. 1
• B. -3
• C. 2
• D. -2

Hint:

The Remainder Theorem is crucial for problems involving divisibility of polynomials. If a polynomial $P(x)$ is divisible by $(x-a)$, then $P(a)$ must be 0. Remember to correctly handle the signs when substituting negative values into powers.

Answer 1

The Correct Option is A

Step 1: Understand the Remainder Theorem.
The Remainder Theorem states that if a polynomial $P(x)$ is divided by $(x-a)$, then the remainder is $P(a)$.
In this problem, the polynomial is $P(x) = x^{140} + 2x^{151} + K$.
It is given that $P(x)$ is divisible by $(x+1)$. This means that when $P(x)$ is divided by $(x+1)$, the remainder is 0.
According to the Remainder Theorem, if $P(x)$ is divisible by $(x+1)$, then $P(-1) = 0$. (Here, $x-a = x+1$, so $a = -1$).
Step 2: Substitute $x = -1$ into the polynomial.
Substitute $x = -1$ into the expression for $P(x)$:
$P(-1) = (-1)^{140} + 2(-1)^{151} + K$
Step 3: Evaluate the powers of -1.
Recall that $(-1)^{\text{even number}} = 1$ and $(-1)^{\text{odd number}} = -1$.
Here, 140 is an even number, so $(-1)^{140} = 1$.
And 151 is an odd number, so $(-1)^{151} = -1$.
Step 4: Substitute the evaluated powers back into the expression for $P(-1)$.
$$P(-1) = 1 + 2(-1) + K$$$$P(-1) = 1 - 2 + K$$$$P(-1) = -1 + K$$ Step 5: Set $P(-1)$ to 0 and solve for K.
Since $P(x)$ is divisible by $(x+1)$, we know that $P(-1) = 0$. So, set the expression for $P(-1)$ equal to 0: $$-1 + K = 0$$Add 1 to both sides:$$K = 1$$ Step 6: Final Answer.
The value of K is 1. $$(1) 1$$