Question

Divide and express the result in the form
P(x) = D(x)Q(X) + R(X)

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

Answer 1

Basically what you have to do is this:

Answer 2

Start with this:
4x^2(x + 4)

You'll end up with
4x^3 + 16x^2

Which means since you have 16x^2 you need to subtract 14x^2 to compensate since 16x^2 - 14x^2 = 2x.
So next you'll do this:

4x^2(x + 4) - 14x^2(x + 4)

since -14x(x + 4) = -14x^2 - 56x

You'll need to add 59x to compensate since -56x + 59x = 3x. Then you'll have

4x^2(x + 4) - 14x^2(x + 4) + 59(x + 4)

59(x + 4) = 59x + 236

But now you need to subtract 232 to compensate since you want 236 - 232 = 4

So finally you'll have
4x^2(x + 4) - 14x^2(x + 4) + 59(x + 4) - 232

And then factor the x + 4 to get

(4x^2 - 14x^2 + 59)(x + 4) - 232

Answer 3

Good luck understanding all of that.

Answer 4

thanks for the help!

Answer 5

can i ask another question?

Answer 6

here it is:

one root of the equation x^3+x^2-2 = 0 is 1. what are the other two roots?

aka how do i find these other two roots?
thanks

Answer 7

If 1 is a root of x^3 + x^2 - 2, then x = 1 is a solution and x - 1 = 0 is a factor.

Answer 8

so generaly, what are the baseline rules or steps that i need to follow to find roots in this type of problem?

Answer 9

im not sure if i understand what you just sent ^

Answer 10

If you have a cubic equation and you know one of the roots, you can reduce the cubic to a quadratic equation. And all quadratic equations are factorable.

Answer 11

Since x - 1 is a factor, re-write x^3 + x^2 - 2 as follows:

x^3 - x^2 + 2x^2 - 2x + 2x - 2 = x^2(x - 1) + 2x(x - 1) + 2(x - 1)
= (x^2 + 2x + 2)(x - 1)

Answer 12

thanks,
how did you get from
x^3 + x^2 - 2
to
x^3 - x^2 + 2x^2 - 2x + 2x - 2

Answer 13

Check your messages.

Answer 14

Well, put it this way

x^2 = -x^2 + 2x^2
0 = -2x + 2x

That probably still doesn't help you understand it though. But I can explain it to you in vyew.

Answer 15

yes, please. it would really help if you explained it to me in vyew. thanks. ill check my messages