# How do I find k if x^4-kx^3-2x^2+x+4 is divided by x-3 and the remainder is 16

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### 2 Answers

Let f(x) = x^4-kx^3-2x^2+x+4

If f(x) is divided by x-3 , then the remainder is 16.

Then we could write:

f(x) = (x-3) * R(x) + 16

Now let us substitute with x =3:

==> f(3) = 0 * R(3) + 16 = 16

==> f(3) = 16 .......(1)

Now let us substitute x=3 in f(x):

f(3) = 3^4 - k* 3^3 -2*3^2 + 3 + 4 = 16

==> 81 - 27k - 18 + 7 = 16

==> -27k = -54

**==> k = 2**

**==> f(x) = x^4 -2x^3 -2x^2 + x + 4 **

**==> f(x) = (x-3)( x^3 + x^2 + x + 4) + 16**

If x^4-kx^3-2x^2+x+4 is divide by x-3 , we get a remainder 16, then

x^3-kx^3-2x^2+x+4-16 = x^4-kx^3-2x^2+x -12 is perfectly divisible by x-3.

We actually divide :

x-3) x^4-kx^3-2x^2 +x -12( x^3

x^4 -3x^3

----------------------

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

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

-----------------------------------------------

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

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

-----------------------------------------------------------

x-3) (-9k +22)x -12 ( (-9k+22)

(-9k +22)x -3 ( -9k +22)

-----------------------------------------------

0

Therefore -12 +3(-9k+22) = 0

-27k + 54 = 0.

-27k = -54.

k = -54/-27 = 2

or k = 2.

Therefore k = 2 for which x^4-kx^3 -2x^2+x+4 diveided by x-3 gives a remainder of 16.