Question

if the remainder on dividing the polynomials 2x ^4-kx^2+5x - 3k+3 by (x+2) is 4,then find the value of k. ​

Answer 1

Question:

• If the remainder on dividing the polynomial 2x⁴ - kx² + 5x - 3k + 3 by ( x + 2 ) is 4, find the value of k.

Given:

• p(x) = 2x⁴ - kx² + 5x - 3k + 3
• Divisor = ( x + 2 )
• Remainder = 4

To find:

• The value of k.

Solution:

Here, p(x) = 2x⁴ - kx² + 5x - 3k + 3

Now, x + 2 = 0

=> x = (-2)

By using remainder theorem,

p(-2) = 2(-2)⁴ - k(-2)² + 5(-2) - 3k+ 3

= 32 - 4k - 10 - 3k + 3

= 32 - 10 + 3 - 3k - 4k

= 25 - 7k .......(I)

So here we get the remainder as 25 - 7k.

But in the question it is given that remainder is 4, therefore both are equal.

So, 25 - 7k = 4

( Both are same remainder)

=> (-7k) = 4 - 25

=> (-7k) = (-21)

=> 7k = 21

( Negative signs got cancelled)

=> k = 21/7

=> k = 3

Answer:

• Therefore the value of k is 3.

Answer 2

Given :-

• Dividend = 2x⁴ - kx² + 5x - 3k + 3

• Divisor = x + 2

• Remainder = 4

To FinD :-

• The value of k is ?

Solution :-

$: \implies \: x + 2 = 0\\ \\ : \implies \: x = - 2$

The polynomial -

$p(x) = 2 {x}^{4} - k {x}^{2} + 5x - 3k + 3$

Using the remainder theo

Putting the x = -2 in the given polynomial

$remainder = 2 \times ( { - 2)}^{4} - k( { - 2)}^{2} + 5 \times ( - 2) - 3k + 3 \\ \\: \implies 4= 2 \times 16 - k \times 4 - 10 - 3k + 3 \\ \\ : \implies 4= 32 - 4k - 10 - 3k + 3 \\ \\ : \implies4 = - 7k + 25 \\ \\ : \implies 7k = 25 - 4 \\ \\ : \implies7k = 21 \\ \\ : \implies \: k = \frac{21}{7} \\ \\ : \implies \: \boxed{ \: k = 3}$