Question

If the remainder on division of x^3 - kx^2 + 13x - 21 by 2x - 1 is -21. Find quotient and there value of k. Hence, find the zeroes of the cubic polynomial​

Answer 1

SOLUTION:-

Let p(x)=x³ -kx² +13x -21 & g(x)= 2x-1

Equating g(x)= 0, we get,

=) 2x-1= 0

=) 2x= 1

=) x= 1/2

Substituting this value of x in p(x) & equating it to the remainder by the remainder theorem, we get:

p(x)= (-21)

$= > {x}^{3} - k {x}^{2} + 13x - 21 = ( - 21) \\ = > ( \frac{1}{2}) {}^{3} - k( \frac{1}{2} ) {}^{2} + 13( \frac{1}{2} ) - 21 = ( - 21) \\ \ \\ = > \frac{1}{8} - \frac{k}{4} + \frac{13}{2} = - 21 + 21 \\ \\ = > \frac{1 - 2k + 52}{8} = 0 \: \: \: \: \: [Take \: L.C.M.]\\ \\ = > 1 - 2k + 52 = 0 \\ \\ = > - 2k + 53 = 0 \\ \\ = > - 2k = - 53 \\ \\ = > k = \frac{53}{2}$

Hope it helps ☺️

Answer 2

heyaa

here's the answer

refer to the attachment

:))