Question

If f(x)(2x^3+x^2+bx-6) leaves the remainder 36 ,when divided by (x-3),find the value of b.With the value of b,factorise f (x)

Answer 1

Hii dear !
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f(x) = 2 x³ + x² + bx - 6 

as (x-3) is a factor so , x = 3 

f(3) = 2 × 3³ + 3² + 3b - 6 

      = 54 + 9 + 3b - 6 
  
      = 63 - 6 + 3b 
      
      = 57 + 3b 
    
      = 3 ( 19 + b ) 
remainder  = 36 

so, 
3 ( 19 + b ) = 36 
19 + b        = 12 
        b        = 12 - 19 
        b        = -7 

Now polynomial changed to , 

2 x ³ + x² - 7x - 6 

2x³ + 2x² - x² - x - 6x - 6

2x² (x + 1) - x(x + 1) - 6(x + 1)

(2x² - x - 6) (x +1)

[2x² - 4x + 3x - 6] (x +1)

(2x +3) (x -2) (x +1)

Answer 2

f(x) = 2x³ + x² + bx - 6 Leave the remainder 36 when put x = 3
f(3) = 2(3)³ + (3)² + b(3) -6
36 = 54 + 9 + 3b - 6
-21 = 3b
-7 = b

now,
f(x) = 2x³ + x² -7x - 6
= 2x³ + 2x² - x² - x - 6x - 6
= 2x²(x + 1) -x(x + 1) - 6(x + 1)
= (2x² - x - 6)(x +1)
={ 2x² - 4x + 3x - 6}(x +1)
= (2x +3)(x -2)(x +1)