when 2x^3 - 3x^2 - 2x + a is divided by ( x-1 ) the remainder is 24. find the value of a

Question

it is a math problem of remainder theorem can u tell me the solution :]

Answer 1

f(1) = 2 - 3 - 2 + a = 24 = remainder by remainder theorem.

 

a = 27

Answer 2

Hi Rodela;

There are two ways to resolve this...

2x^3 - 3x^2 - 2x + a is divided by ( x-1 )

We can divide this into two binomial expressions...

For the FOIL...

FIRST must be either (2x²)(x) or (2x)(x²)=2x³

OUTER must be (-1)(2x)=-2x

INNER must be (-3)(x²)=-3x²

LAST must be (-3)(-1)=3

2x³-3x²-2x+3=(2x-3)(x²-1)

x²-1=(x+1)(x-1)

2x³-3x²-2x+3=(2x-3)(x+1)(x-1)

The constant is 3.

We want a remainder of 24.

3+24=27

2x³-3x²-2x+3+24

2x³-3x²-2x+27

a=27

 

We can also factor this into a binomial and trinomial.

2x^3 - 3x^2 - 2x + a

(x-1)(...

FIRST...(x)(2x²)=2x³

(x-1)(2x²....

SECOND...(-1)(2x²)=-2x² and (x)(-x)=-x², -2x²-x²=-3x²

(x-1)(2x²-x...

THIRD...(-1)(-x)=x and (x)(-3)=-3x, x-3x=-2x

(x-1)(2x²-x-3)

Let's do the multiplication...

x(2x²-x-3)=2x³-x²-3x

-1(2x²-x-3)=    -2x+x+3

(x-1)(2x²-x-3)=(2x³-3x²-2x+3)

We need a remainder of 24.

We have a constant of 3.

a=24+3=27

a=27

2x³-3x²-2x+3+24

2x³-3x²-2x+27