This one is a doozy?

Question

(6x^4+17x^3+16x^2+14x+12) divided by 2x+3

I have attempted to find the following:
to solve Q(x) + R(x)/ 2x+3

so far:
6x^4+17x^3+16x^2+14x+12

how do I get the above substitution?

divided by (2x+3)

Answer 1

An easy way to do this is by long division, surprisingly. So, by letting "I" be the same as a division house, or "goes into"...

(2x + 3) I (6x^4+17x^3+16x^2+14x+12)

We see that we need a 3x^3 to make
2x*3x^3 = 6x^4

This will give us our first term in Q(x) and we will have
3x^3*(2x + 3) = 6x^4 + 9x^3
which we will call P1(x)

Now subtract P1(x) from the denominator:
(6x^4+17x^3+16x^2+14x+12) - (6x^4 + 9x^3) = 8x^3+16x^2+14x+12

Now we've eliminated the first term in our large denominator.
Next we need a term (y) that will allow 2x * y = 8x^3
Thus "y" is 4x^2, which is now going to be our second term in Q(x).

Repeating the same process above:
4x^2*(2x+3) = 8x^3 + 12x^2, which we will now call P2(x)

Subtract P2(x) from our remaining denominator:
(8x^3+16x^2+14x+12) - (8x^3 + 12x^2) = 4x^2+14x+12

now you need another term (y) to make 2x * y = 4x^2
Thus y is now 2x.

Continue this process until you complete Q(x) and anything remaining will be your R(x)

I found Q(x) = 3x^3+4x^2+2x+4
and R(x) = 0

Answer 2

3x^3 + 4x^2 + 2x + 4
2x+3\6x^4+17x^3+16x^2+14x+12
-(6x^4+ 9x^3)
8x^3+16x^2
-(8x^3+12x^2)
4x^2+14x
-(4x^2 + 6x)
8x +12
-(8x + 12)

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

Answer 3

use long division
6x^4 / 2x = 3x^3
(2x+3) * 3x^3 = 6x^4 + 9x^3
and so on thru rest of the expression
answer is 3X^3 + 4x^2 +2x + 4

Answer 4

3x^3 + 4x² + 2x + 4

no remainder

Answer 5

carefully