Question

when 2x³-5x²+4x-8 is divided by ax²+bx+c the quotient is x-2 and the remainder is bx-a find the value of a,b and c?​

Answer 1

Answers:

a = 2

b = -1

c = 3

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Work Shown:

(numerator)/(denominator) = quotient + (remainder)/(denominator

(2x^3-5x^2+4x-8)/(ax^2+bx+c) = x-2 + (bx-a)/(ax^2+bx+c)

2x^3-5x^2+4x-8 = (x-2)(ax^2+bx+c) + (bx-a)

2x^3-5x^2+4x-8 = x(ax^2+bx+c)-2(ax^2+bx+c) + (bx-a)

2x^3-5x^2+4x-8 = ax^3+bx^2+cx-2ax^2-2bx-2c + bx-a

2x^3-5x^2+4x-8 = ax^3+(bx^2+2ax^2) + (-2bx+cx+bx) + (-a-2c)

2x^3-5x^2+4x-8 = ax^3+(b+2a)x^2 + (-b+c)x + (-a-2c)

The x^3 term of the left hand side (LHS) is 2x^3, while the x^3 term of the right hand side (RHS) is ax^3. This must mean a = 2.

The x^2 term of the LHS is -5x^2. The x^2 term of the RHS is (b+2a)x^2

Plugging in a = 2 leads to (b+2a)x^2 turning into (b+4)x^2. Equating the two x^2 coefficients has us get b+4 = -5 solve to b = -1

Now focus on the x terms. The LHS has 4x and the RHS has (-2b+c)x. Plug in b = -1 to get (-b+c)x turn into (c+1)x. Set this equal to 4x and you should find that c = 3

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In summary, we have:

a = 2

b = -1

c = 3

Note how the last term on the RHS is -a-2c = -2-2(3) = -8 which matches with the last term on the LHS. This helps confirm we have the correct values for a,b,c. Further confirmation would involve polynomial long division.

Answer 2

a = 2; b = -1; and c = 3

Explanation:

If the quotient of
`2x^3-5x^2+4x-8` divided by
`ax^2+bx+c`, gives
`x-2` , with a remainder of
`bx-a`, then the product (
`ax^2+bx+c`, ) * (
`x-2` ) plus
`bx-a` should equal:
`2x^3-5x^2+4x-8`

And we can investigate the values of a, b, and c via operating the product and afterwards identifying like terms as shown below:

`(x-2) * (ax^2+bx + c)+bx-a= ax^3 +bx^2+cx-2ax^2-2bx-2c+bx-a=\\=ax^3+(b-2a)x^2+(c-2b+b)x-2c-a=2x^3-5x^2+4x-8`

And therefore, making the coefficients of like terms equal, we get:

`a = 2\\b-2a=-5;\,\,and\,\, b=-1\\c-b=4;\,\,and\,c=3\\-2c-a=-6-2=-8`