Question

Let R1 and R2 be the remainders when polynomial kx3+3x2-14x+5 and
x3-5x2+kx+12 are divided by x+3 and x-2 respectively. If R1- R2= 16 , find
the value of k.
Answer fast

Answer 1

Answer:

$k = 2$

Step-by-step explanation:

Let $f(x)=kx^{3} + 3x^{2} - 14x + 5$ and  $p(x) = x^{3} - 5x^{2} + kx + 12$

The question states that:-

$i)$   When $kx^{3} + 3x^{2} - 14x + 5$ is divided by $x + 3$ it gives a remainder $R_1$

$ii)$  When $x^{3} - 5x^{2} + kx + 12$ is divided by $x-2$ it gives a remainder $R_2$

$iii)$ $R_1 - R_2 = 16$

⇒ $f(-3) = R_1$ and $p(2) = R_2$

⇒ $f(-3)= k(-3)^{3} + 3(-3)^{2} -14(-3) + 5 = R_1$

⇒ $-27k + 27 + 42 + 5 = R_1$

⇒ $-27k + 74 = R_1$

⇒ $p(2) = (2)^{3} - 5(2)^{2} + k(2) + 12 = R_2$

⇒ $8 - 20 + 2k + 12 = R_2$

⇒ $2k = R_2$

We know that $R_1 - R_2 = 16$

⇒ $-27k + 74 - 2k = 16$

⇒ $-29k = 16 - 74$

⇒ $-29k = -58$

⇒ $k = 2$