Question

The remainder are same where 3x^4 - 4x^2 + ke - 5 is divided by (x - 3) and (x + 2). Find the value of k.

Answer 1

f(x) = 3x⁴-4x²+kx-5
x-3 => x= 3
f(3) = 3(3)⁴-4(3)²+k(3) = 5
= 243 - 36 + 3k = 5
= 207 - 5 = -3k
= 202 = -3k
=> k = -202/3 = -68

Answer 2

Heya here .

Solution Given here ⬇⬇⬇
$sol. \\ \\ x - 3 = 0 \\ x = 3 \\ p(x) = 3x ^{4}- 4 {x}^{2} + kx - 5 \\ p(3) = 3(3 {)}^{4} - 4(3 {)}^{2} + 3k - 5 \\ = > 81 - 108 + 3k - 5 \\ = > 3k - 32 \\ \\ x + 2 = 0 \\ x = - 2 \\ p( - 2) = 3( - 2 {)}^{4} - 4( - 2 {)}^{2} + k(2) - 5 \\ = > - 24 - 16 + 2k - 5 \\ = > 2k - 35 \\ \\ 3k - 32 = 2k - 35 \\ 3k - 2k = - 35 + 32 \\ k = - 3$
THEREFORE , the value of k is -3
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