Math, asked by naredalakhan, 8 months ago

The expression 2x^3 + ax^2 + bx-2 leaves remainder 7 and 0 when divided by 2x-3 and x+2 respectively.Calculate the values of a and b.​

Answers

Answered by Revou123
2

Answer:

a = -3/5 and b = -39/5

Step-by-step explanation:

(i) 2x-3 = 0  

x = 3/2

by using remainder thearom

put the value of x in  f(x) 2x^3 + ax^2 + bx-2

f(3/2) = 2 ×( 3/2)^3  + a × (3/2)^2 + b × 3/2 -2 = 7

        = 2 × 27/8 +  a × 9/4 + 3b/2 - 2 = 7

        = 27/4 + 9a/4 + 3b/2 = 7 -2

        = (9a + 6b)/4 = 5 - 27/4 = 9a + 6b = -7        

        = 3a + 2b = -21                                                                 ---(1)

(ii) x + 2 = 0

x = -2

by using remainder thearom

put the value of x in  f(x) 2x^3 + ax^2 + bx-2

f(-2)  = 2(-2)^3 + a(-2)^2 + (-2b) -2 = 0

      = -16 + 4a - 2b - 2 = 0

       = 4a - 2b = 18

        = 2a - b = 9                                                                  -----(2)

solving eq. 1 and 2 we get the value of

a = -3/5 and b = -39/5

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