find the value of a and b so that 2×3+ax2+x+b has x+2 and 2x-1 as factor​

2 thoughts on “find the value of a and b so that 2×3+ax2+x+b has x+2 and 2x-1 as factor​”

1. Answer:

   f(x)=2×3+ax2−bx+3

   At x=2

   f(2)=15

   f(1)=0

   f(x)=2×3+ax2−bx+3

   f(1)= 2+a-b+3=0

   a-b+5=0 ____A

   f(x)= 2×3+ax2−bx+3

   f(2)=2(23)+a(22)−2b+3=15

   4a-2b=-4

   Multiply A by 2 and subtract from above equation

   4a-2b=-4

   2a-2b+10=0

   2a-10=-4

   2a= 6

   a=3

   From A

   3-b+5=0

   8-b=0

   b=8

   So a=3 and b=8

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2. Therefore the value of a = 5 and b = -2.

   Step-by-step explanation:

   Given f(x) = 2x^3 + ax^2 + x + b.

   Given that 2x – 1 and x + 2 are the factors of f(x).

   => 2x – 1 = 0

   => x = 1/2

   Plug x = 1/2 in f(x), we get

   => 2(1/2)^3 + a(1/2)^2 + (1/2) + b = 0

   => 1/4 + a/4+1/2+b=0

   => 3/4 + a/4 + b = 0

   => a+ 3+ 4b = 0

   => a + 4b = -3 (1)

   When x + 2:

   => x + 2 = 0

   => x = -2.

   Plug x = -2 in f(x), we get

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

   => -16 + 4a – 2 + b = 0

   => 4a + b – 18 = 0

   => 4a + b = 18 (2)

   On solving (1) * 4 & (2), we get

   4a +16b-12

   4a + b = 18

   15b = -30

   b = -30/15.

   b = -2

   Substitute b = -2in (1), we get

   => a + 4b = -3

   => a + 4(-2) = -3

   => a – 8 = -3

   => a = -3 + 8

   a = 5.

   Therefore the value of a = 5 and b = -2.

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