Question

Determine the value of k if f(x) = 2x^4 +3x^3 +2kx^2 +3x +6 is exactly divisible by (x+1)




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Answer 1

Answer:

Polynomial2x

4

+3x

3

+2kx

2

+3x+6

Since \: x + 1 \: divides \: the \: polynomial \: completelySincex+1dividesthepolynomialcompletely

So,So,

p(x)\:=\:x+1p(x)=x+1

p(x)\:=\:0p(x)=0

So,\:x + 1 = 0So,x+1=0

x = - 1x=−1

Putting \: the \: value \: of \: x \: :-Puttingthevalueofx:−

2 {x}^{4} + 3 {x}^{3} + 2k {x}^{2} + 3x + 62x

4

+3x

3

+2kx

2

+3x+6

2 \times { - 1}^{4} + 3 \times { - 1}^{3} + 2 \times k \times { - 1}^{2} + 3 \times - 1 + 6 = 02×−1

4

+3×−1

3

+2×k×−1

2

+3×−1+6=0

2 \times 1 + 3 \times - 1 +2k + ( - 3) + 6 = 02×1+3×−1+2k+(−3)+6=0

2 - 3+ 2k - 3 + 6 = 02−3+2k−3+6=0

- 1 + 2k - 3 + 6 = 0−1+2k−3+6=0

- 1 + 2k - 3 = 0−1+2k−3=0

- 1 + 2k = 0 + 3−1+2k=0+3

2k = 1 + 32k=1+3

k = \frac{4}{2}k=

2

4

\therefore \: k = 2∴k=2

hope it help u bro

Answer 2

Here, zero of x + 1:-

x + 1 = 0

=> x = - 1

∵ f(x) = 2x⁴ + 3x³ + 2kx² + 3x + 6

=> f(- 1)² = 0

[As (x - 1) completely divides f(x).]

=> 2(- 1)⁴ + 3(- 1)³ + 2k(- 1)² + 3(- 1) + 6 = 0

=> 2 - 3 + 2k - 3 + 6 = 0

=> 2k + 2 = 0

=> 2k = - 2

=> k = - 1.

Aliter:

(See attachment.)

R = (2k - 1) + 3x + 6 = 0

=> 2k + 3x + 5 = 0

=> 2k + 3x = - 5

=> 3x + 2k = - 3 - 2

=> 3x + 2k = 3(- 1) + 2(- 1).

Therefore, by comparing, k = - 1.