Question

Find the value of k so that (x^4 - 2x^2 + kx + 6) is divided by (x - 2), the remainder is 0.

Answer 1

We want to calculate the following division

`(x^4-2x^2+kx+6)/(x-2)`

To efetuate this divison, we start by dividing the leading term of the dividend by the leading term of the divisor:

`(x^4)/(x)=x^3`

Then, we multiply this result by the divisor:

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

Then, subtract the dividend from the obtained result:

`(x^4-2x^2+kx+6)-(x^4-2x^3)=2x^3-2x^2+kx+6`

Then, we can rewrite our division as:

`(x^(4)-2x^(2)+kx+6)/(x-2)=x^3+(2x^3-2x^2+kx+6)/(x-2)`

The remainder of this first division still have a polynomial on the dividend, therefore, we can iterate the previous process until we find the final result of this division.

`\begin{gathered} (x^(4)-2x^(2)+kx+6)/(x-2)= x^(3)+(2x^(3)-2x^(2)+kx+6)/(x-2) \\ =x^3+2x^2+(2x^2+kx+6)/(x-2) \\ =x^3+2x^2+2x+(x(k+4)+6)/(x-2) \\ =x^3+2x^2+2x+k+4+(2k+14)/(x-2) \end{gathered}`

And this is the final result of the division.

`(x^4-2x^2+kx+6)/(x-2)=x^3+2x^2+2x+k+4+(2k+14)/(x-2)`

We want to find the value of k that will make the remainder equal to

The remainder of this division is

`(2k+14)/(x-2)`

Then, we just need to solve for k the following equation

`\begin{gathered} (2k+14)/(x-2)=0\implies2k+14=0 \\ 2k=-14 \\ k=-(14)/(2) \\ k=-7 \end{gathered}`

The value of k must be - 7.

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