Question

What must be subtracted from 4x^4 + 2X^3 - 8X^2 + 3X - 7 so that it may be exactly divisible by 2x^2 + x - 2 ?

Answer 1

First multiply 2x^2 + x - 2 by 2x^2:

2x^2(2x^2 + x - 2)

Answer 2

Use polynomial long division to divide the polynomials. The answer is the remainder.

Answer 3

thanku

Answer 4

yw

Answer 5

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

4x^4 + 2X^3 - 8X^2 + 3X - 7
-4x^4 - 2x^3 + 4x^2

You subtract this: -4x^2 + 3x - 7

Because if you do that:

4x^4 + 2x^3 - 8x^2 + 3x - 7 - (-4x^2 + 3x - 7)

= 4x^4 + 2x^3 - 8x^2 + 4x^2 + 3x - 3x + 7 - 7
= 4x^4 + 2x^3 - 4x^2

Which is divisible by 2x^2 + x - 2

Answer 6

The way this problem is worded, there is not a unique solution.
Let me explain what I mean with numbers, which we are more familiar with.
We know that 57 is not divisible by 8.

Let's say I ask the question "what must be subtracted from 57 to make it divisible by 8?"

My approach was:
Divide 57 by 8 to get a remainder of 1.
Then subtract 1 from 57 to get 56 which is divisible by 8.
My answer, then is 1, the remainder of the division.

Another approach is:
Multiply 8 by 2 to get 16. 16 is definitely a multiple of 8 and obviously divisible by 8.
Then since 57 - 16 = 41, if you subtract 41 from 57 you get 16 which is divisible by 8.
This was @Hero's approach.

@Hero and I have two correct answers to the problem.
By multiplying 2x^2 + x - 2 by 2x^2, Hero came up with a multiple of 2x^2 + x - 2 which is obviously divisible by 2x^2 + x - 2 because that is what a multiple is.

My approach was to perform polynomial long division and come up with a remainder. Then if the remainder is subtracted from the dividend, then resulting polynomial is divisible by the divisor.