Question

The polynomial
$kx {}^{4} + 3x {}^{3} + 6$
when divided by x- 2 leaves a remainder which is double the reminder left by the polynomial
$2x {}^{3} + 17x + k$
when divided by x-2 . Find the value of k.​

Answer 1

Answer :-

Value of k is 5.

Explanation :-

Let p(x) = kx^4 + 3x³ + 6

f(x) = 2x³ + 17x + k

Given

When p(x) is divided by (x - 2) leaves a remainder which is doble the remainder left by the f(x) when divided by (x - 2)

Finding the zero of (x - 2)

x - 2 = 0

x = 2

By remainder theorem, remainders are p(2) and f(2) respectively.

According to the question

⇒ p(2) = 2{ f(2) }

⇒ k(2)^4 + 3(2)³ + 6 = 2{ 2(2)³ + 17(2) + k}

⇒ k(16) + 3(8) + 6 = 2{ 2(8) + 34 + k}

⇒ 16k + 24 + 6 = 2(16 + 34 + k)

⇒ 16k + 30 = 2(50 + k)

⇒ (16k + 30)/2 = 50 + k

⇒ (16k/2) + (30/2) = 50 + k

⇒ 8k + 15 = 50 + k

⇒ 8k - k = 50 - 15

⇒ 7k = 35

⇒ k = 35/7

⇒ k = 5

∴ the value of k is 5.

Answer 2

Answer:

$\huge\underline\mathfrak{answer:-}$

$let \: \: f(x) =k {x}^{4} + 3 {x}^{3} + 6 \\ \\ and \: \\ \\ g(x) = 2 {x}^{3} + 17x + k \\ \\ according \: to \: question \\ x - 2 = 0 \implies \: x = 2 \: for \: f(x) \\ \\ and \: g(x) \\ \\ hence \: according \: to \: q. \\ \\ f(2) = 2g(2) \\ \\ put \: x = 2 \: \\ \\ \implies 16k + 24 + 6 = 2(16 + 34 + k) \\ \\ \implies \: 16k + 30 = 100 + 2k \\ \\ \implies \: 14k = 70 \\ \\ \implies \red{ \boxed{k = 5} }\\ \\ \huge \boxed{ \mathfrak{answer}}$