Verified answer

1) 3x^3-x-5 / x-2

Can't be done. I was hoping (like others) that there was a typo involved.....that you meant 3x^2 but you repeated 3x^3 so not possible. Using long division I got 3x^2 + 6x + 11 but a remainder of 17.

2) 2x^2+9x+4 / 2x+1........factor numerator

(2x + 1)(x + 4)/(2x + 1).......2x + 1 cancels

x + 4 ..........answer

3) x^2+5x+6 / x+2 .........factor numerator

(x + 3)(x + 2)/(x + 2).......x + 2 cancels

x + 3...........answer

Let's try the first one. If you divide an ordinary number, let's call it N, by a divisor D, you want to get a number Q so that N/D=Q. Turn that around and you see that D*Q=N. Got it? Right. It works the same for polynomials, so if 2x^2+9x+4 / 2x+1=Q, then Q*(2x+1)=2x^2+9x+4. The difference is that Q is now a polynomial too. Let's write it out with symbols: Q=ax+b. Now we need to figure out what a and b are. To do that, we go back to Q*(2x+1)=2x^2+9x+4 then replace Q by its expansion: (ax+b)(2x+1)=2x^2+9x+4. Now we multiply the parentheses on the left side: 2ax^2+(a+2b)x+b=2x^2+9x+4. The values of a and b can now be read from terms of the same order. Taking x^2 you get 2a=2, so a=1. The next term gives you 1+2b=9, and the last one lets you check your calculation. Your answer for the second problem is therefore x+3.

I took the second problem as an example because the first polynomial is of third degree. That means your Q for the first problem will be of degree 2 and the equations will be a bit longer.

Is there a typing error in 3x^3-x-5 / x-2 ?

1. (3x + 5)(x - 2)

answer (3x + 5)

2. (2x + 1)(x + 4)

answer(x + 4)

3. (x + 3)(x + 2)

answer(x + 3)

1.)

Divide by 2

2

Enter coefficients when there is no value(as for x^2 use 0)

6 below should be below the 0, 12 below -1 and 22 below -5

3 0 -1 -5

6 12 22

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3 6 11 17

answer

3x^2 + 6x + 11 remainder 17