Question

19. The polynomial p(x) = 2x^3 + kx^2 – 3x + 5 and q(x) = x^3 + 2x^3 - x + k, when divided by (x - 2)
leave the same remainder r1 and r2 respectively. Find the value of k, if r1 - 12 = 0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:p(x) = {2x}^{3} + {kx}^{2} - 3x + 5$

and

$\rm :\longmapsto\:q(x) = {x}^{3} + {2x}^{3} - x + k = {3x}^{3} - x + k$

Case :- 1

Given that when

$\rm :\longmapsto\:p(x) \: is \: divided \: by \: (x - 2), \: it \: gives \: remainder \: r_1$

We know,

Remainder Theorem states that if a polynomial f(x) is divided by linear polynomial (x - a), then remainder is f(a).

So,

$\rm :\longmapsto\:r_1 = p(2)$

$\rm :\longmapsto\: r_1= 2{(2)}^{3} + {k(2)}^{2} - 3(2) + 5$

$\rm :\longmapsto\: r_1= 16 + 4k -6 + 5$

$\rm :\longmapsto\: r_1= 15 + 4k - - - (1)$

Case :- 2

Given that when

$\rm :\longmapsto\:q(x) \: is \: divided \: by \: (x - 2), \: it \: gives \: remainder \: r_2$

So, By Remainder Theorem,

$\rm :\longmapsto\:r_2 = q(2)$

$\rm :\longmapsto\:r_2 = {3(2)}^{3} - 2 + k$

$\rm :\longmapsto\:r_2 = 24- 2 + k$

$\rm :\longmapsto\:r_2 = 22 + k - - - (2)$

Now,

According to statement,

$\rm :\longmapsto\:r_1 - r_2 = 0$

$\rm :\longmapsto\:r_1 = r_2$

$\rm :\longmapsto\:15 + 4k = 22 + k$

$\rm :\longmapsto\:4k - k= 22 - 15$

$\rm :\longmapsto\:3k= 7$

$\bf\implies \:k = \dfrac{7}{3}$

Additional Information :-

Factor theorem :-

This theorem states that if a polynomial f(x) is divided by linear polynomial (x - a), then f(a) = 0