Question

If the following polynomials, when p(x) and q(x) are divided by (x-4), leave the same
reminder in each case. Find the value of'k':*

p (x)
kxcube + 3x2 - 3 & q(x) = 2x3– 5x + k​

Answer 1

Solution :-

→ p(x) = kx³ + 3x² - 3

→ p(4) = k(4)³ + 3(4)² - 3

→ p(4) = (64k) + 3*16 - 3

→ p(4) = (64k) + 48 - 3

→ p(4) = (64k) + 45

Similarly,

→ q(x) = 2x³ - 5x + k

→ q(4) = 2(4)³ - 5*4 + k

→ q(4) = 2*64 - 20 + k

→ q(4) = 128 - 20 + k

→ q(4) = 108 + k

Now, since remainder is same in each case,

→ 64k + 45 = 108 + k

→ 64k - k = 108 - 45

→ 63k = 63

→ k = 1 (Ans.)

Hence, value of k will be 1 .

Answer 2

Solution :

The polynomials p(x) = kx³ + 3x² -3 & q(x) = 2x³ - 5x + k are both divided by x-4;

x-4 = 0

x = 4

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

$\mapsto\sf{p(x) = q(x) }\\\\\mapsto\sf{p(4) = q(4)}\\\\\mapsto\sf{k(4)^{3} + 3(4)^{2} -3 = 2(4)^{3} -5(4) +k}\\\\\mapsto\sf{k(64) +3(16) - 3=2\times 64 -20 + k}\\\\\mapsto\sf{64k + 48 - 3=128 -20+k}\\\\\mapsto\sf{64k + 45 = 108 + k}\\\\\mapsto\sf{64k -k=108 -45}\\\\\mapsto\sf{63k = 63}\\\\\mapsto\sf{k=\cancel{63/63}}\\\\\mapsto\bf{k=1}$

Thus;

The value of k will be 1 .