Question

find the value of 'm' so that (2x-3) is a factor 6x3 -x3-10x+m

Answer 1

First of all we have to find the zero of the divisor

so zero of divisor is

2x - 3 = 0

=> 2x = 3

=> x = 3/2

If (2x - 3) is a factor of (6x³ - x² - 10x + m) then by remainder theorem, remainder will be zero.

=> Putting the value of x we get

6(3/2)³ - (3/2)² - 10(3/2) + m = 0

=> 81/4 - 9/4 - 15 + m = 0

=> 81/4 - 9/4 - 60/4 + m = 0

=> 12/4 + m = 0

=> 3 + m = 0

=> m = (-3)


Hope it helps dear friend ☺️✌️

Answer 2

$\textbf{ANSWER:}$

2x - 3 = 0

=》 2x = 3

=》 x = 3/2 ...(1)

p(x) = 6x^3 - x^2 - 10x + m ...(2)

Putting (1) in (2):

$6 \times ( { \frac{3}{2} })^{3} - ({ \frac{3}{2} })^{2} - 10\times{\frac{3}{2}}+ m = 0\\ \\ = > 6 \times \frac{27}{8} - \frac{9}{4} - 15 + m = 0\\ \\ = > \frac{162}{8} - \frac{9}{4} = 15 - m \\ \\ = > \frac{144}{8} = 15- m \\ \\ = > 144 = 120 - 8m \\ \\ = > 144 - 120 = - 8m \\ \\ = > 24 = - 8m \\ \\ = > \frac{24}{ - 8} = m \\ \\ = > (-3) = m$

That's the answer!