this one needs terms and remainder?

Question

(4x^4+13x^3+6x^2+11x+19) divided by (4x+5)?

Answer 1

Q=x^3+2x^2-x+4
the answerQ=x^3+2x^2-x+4
R=-1

Answer 2

this is a problem that must be solved using long division, factoring would be quite difficult unless you have a pretty powerful calculator (TI-89 for starters).

so

(4x + 5) => 4x^4 + 13x^3 + 6x^2 + 11x + 19


to get rid of the x^4, we'd need to multiply 4x by x^3 (and then the five as well)

(4x^4 + 13x^3 + 6x^2 + 11x + 19) - (4x^4 + 5x^3)

= 8x^3 + 6x^2 + 11x + 19

to get rid of the 8x^3, we need to multiply 4x by 2x^2 (and then the five as well)
(8x^3 + 6x^2 + 11x + 19) - (8x^3 + 10x^2)

= -4x^2 + 11x + 19

to get rid of -4x^2, we need to multiply 4x by -1x (and then the five as well)

(-4x^2 + 11x + 19) - (-4x^2 -5x)

= 16x+ 19

to get rid of the 16x, multiply 4x by 4 (and then to the five as well)

(16x+ 19) - (16x + 20)

= -1

so the final answer is:

x^3 + 2x^2 -x +4 [R-1]

Answer 3

This was just asked and answered, see previous posts...