Consider the polynomial $$ P(x)=k x^{3}-3 x^{2}+4 x+3 $$. Find the value of $$ k $$ such that the remainder is 7 when $$ P(x) $$ is divided by $$ x-2 $$. $$ k=\square $$

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We apply the Remainder Theorem, which states that the remainder when a polynomial $$ P(x) $$ is divided by $$ x - a $$ is $$ P(a) $$. For this problem, we have $$ a = 2 $$, so we need to have

$$
P(2) = 7.
$$

Given

$$
P(x) = kx^3 - 3x^2 + 4x + 3,
$$

substitute $$ x = 2 $$:

$$
P(2) = k(2)^3 - 3(2)^2 + 4(2) + 3.
$$

Calculate each term:

$$
2^3 = 8 \quad \text{and} \quad 2^2 = 4.
$$

Hence,

$$
P(2) = 8k - 3(4) + 8 + 3 = 8k - 12 + 8 + 3.
$$

Simplify the constant terms:

$$
-12 + 8 + 3 = -1.
$$

Thus,

$$
P(2) = 8k - 1.
$$

Setting the remainder equal to 7 gives:

$$
8k - 1 = 7.
$$

Now, solve for $$ k $$:

$$
8k = 7 + 1 = 8,
$$

$$
k = \frac{8}{8} = 1.
$$

So, the value of $$ k $$ is $$ 1 $$.
$$ k = 1 $$

Consider the polynomial .
Find the value of such that the remainder is 7 when is divided by .

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Consider the polynomial . Find the value of such that the remainder is 7 when is divided by .