Question

what number must be added to 2x^3-7x^2+2x so that the resulting polynomial leaves the remainder -2 when divided by 2x-3​

Answer 1

Answer:

-4x-2

Step-by-step explanation:

Given polynomial,

p(x)=2x^3-7x^2+2x

g(x)=2x-3

dividing. p(x) by g(x), we get

2x-3)2x^3-7x^2+2x(x^2-2x

2x^3-3x^2

- +

0 -4x^2+2x

-4x^2+6x

+ -

0 -4x

q(x)=x^2-2x

and, r(x)=-4x

Now,

According to division algorithm

p(x)=g(x)*q(x)+r(x)

2x^3-7x^2+2x=(2x-3)(x^2-2x)+(-4x)

adding 4x-2 on both side, we get

2x^3-7x^2+2x+4x-2=(2x-3)(x^2-2x)+(-4x)+4x-2

2x^3-7x^2+6x-2=(2x-3)(x^2-2x)+(-2)

Hence, 4x-2 is added to the given polynomial p(x) to leave the remainder=(-2).

Answer 2

Answer:

-4x-2

Step-by-step explanation: