Math, asked by manav7226, 11 months ago


If (x3 + ax2 + bx +6) has (x - 2) as a factor and leaves a remainder 3 when
divided by (x - 3), find the values of a and b.​

Answers

Answered by jainmiley1234
11

Answer:

Step-by-step explanation:

x^3 + ax^2 +bx + 6

if  x-2 is  a factor  

 then  x=2

 now putting the value of x in eq 1

2^3 + 2^2a +2b +6

4a + 2b + 14       .......(1)

if   x-3 is a factor so x=3

and remainder = 3

3^3 + 3^2a+3b+6 = 3

27+9a+3b+6=3

9a+3b+33=3

9a+3b+30=0         ......(2)

\\now from equation 1 and   2

      4a + 2b + 14

      9a+3b+30=0

⇒    2a + b + 7 = 0

      3a + b+10 = 0

⇒a = -3  

and from eq 1

4a + 2b + 14 =0

4*(-3) + 2b + 14 =0

-12 +2b = -14

2b = -14 + 12

2b = -2

b = -1

Answered by amitnrw
1

Given : x³  +  ax² + bx + 6 has  (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3)

To Find :  values of a and b.

Solution:

x³  +  ax² + bx + 6 has  (x – 2) as a factor

=>for  x = 2  value = 0

=> 2³  +  a.2² + b.2 + 6  = 0

=> 8 + 4a + 2b + 6 = 0

=> 4a + 2b = - 14

=> 2a + b = - 7

leaves a remainder 3 when divided by (x – 3)

=> for x = 3 value  =  3

=> 3³  +  a.3² + b.3 + 6  = 3

=> 27 + 9a + 3b + 6 = 3

=> 9a + 3b = -30

=> 3a + b = -10

3a + b = -10

2a + b = - 7

=> a  = - 3

b  = - 1

Value of a = - 3

Value of b = -1

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