Question

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

Answer 1

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Answer 2

$Polynomial \: 2 {x}^{4} + 3 {x}^{3} + 2k {x}^{2} + 3x + 6$

$Since \: x + 1 \: divides \: the \: polynomial \: completely$

$So,$

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

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

$So,\:x + 1 = 0$

$x = - 1$

$Putting \: the \: value \: of \: x \: :-$

$2 {x}^{4} + 3 {x}^{3} + 2k {x}^{2} + 3x + 6$

$2 \times { - 1}^{4} + 3 \times { - 1}^{3} + 2 \times k \times { - 1}^{2} + 3 \times - 1 + 6 = 0$

$2 \times 1 + 3 \times - 1 +2k + ( - 3) + 6 = 0$

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

$- 1 + 2k - 3 + 6 = 0$

$- 1 + 2k - 3 = 0$

$- 1 + 2k = 0 + 3$

$2k = 1 + 3$

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

$\therefore \: k = 2$