Question

If f(x)=24x^3 + px^2 - 5x+q has two factors (2x+1) and (3x-1),the value of p must be?

Answer 1

f(x) = 24x^3 + px^2 - 5x + q 
factors are 2x + 1 & 3x -1
let us take the factor of 2x + 1 = 0 
2x = - 1 
x = -1/2 
apply the x value in f(x) we get 
f(-1/2) = 24(-1/2)^3 + p(-1/2)^2 - 5(-1/2) + q 
          = 24 (-1/8) + p(1/4) + 5/2 + q 
          = -3 + p(1/4)  + 5/2+q 
          = -3 +5/2+ q + p(1/4) 
          = (-6+5)/2 + q + p (1/4) 
          =  p(1/4) +q -1/2 
          p(1/4) + q = 1/2 
          p + 4q = 4/2 
          p + 4q = 2      consider this as equ (1) 
now take the another factor 3x -1 = 0 
 3x = 1 
x = 1/3 
apply the x value in f(x) we get 
f(1/3) = 24 ( 1/3) ^3 + p(1/3) ^2 - 5(1/3) + q 
        = 8 + p(1/3) - 5/3+q 
       = p(1/3) + q + 8 - 5/3 
       = p (1/3) + q + (24 -5) /3 
       = p(1/3) + q + 19 /3 
     p(1/3) +q = -19/3 
    p + 3q = ((-19) (3) )/3 
   p + 3q = -19 consider it as equ (2) 
solving (1) & (2) we get , 
           p + 4q = 2 
           p + 3q = - 19 (subtract we get ) 
....................................
               q = 21 
apply q value in any one the above equ we get p value 
p + 4( 21) = 2 
p + 84 = 2 
p = 2 -84 
p =-82 
therefore p is - 84 

Answer 2

p = -2, q =1

Step-by-step explanation:

x= -1/2

f(x) =

$24( \frac{ - 1}{2} )^{3} + p ({ \frac{ - 1}{2} )}^{2} - 5( { \frac{ - 1}{2} })^{2} + q$

=>

$- 3 + \frac{p}{4} + \frac{5}{2} + q$

=> -12+p-10+4q=0

=> p+4q=2 - (I)

x=1/3

f(x)=

$24( \frac{1}{3})^{3} + p( \frac{1}{3})^{2} - 5( \frac{1}{3}) + q$

=> 8+p-15-9q=0

=> p-9q=7 - (II)

By solving eq I and II,

p+9q =7

- p+4q =2

_____________

5q =5

q = 1

p = 2-4q (from eq I)

= 2-4

= -2