Question

What is the remainder when (2x4 – 3x3 + x2 – 3x – 72) ÷ (x – 2)

Answer 1

could you maybe clarify your multiplication sign and x

Answer 2

What is the remainder when (2x^4 – 3x^3 + x^2 – 3^x – 72) ÷ (x – 2)

Answer 3

i think for 3^x you meant 3x...?

Answer 4

I think it is this:

\(\dfrac{2x^4-3x^3+x^2-3x-72}{x-2}\)

Answer 5

Do you know how to do a synthetic division?

Answer 6

no

Answer 7

the options are 6 ,3 -6 or-66

Answer 8

It goes like this: take the solution of x-2=0, so that is 2. Next to it, write down all the coefficients of 2x4 – 3x3 + x2 – 3x – 72, that is: 2, -3, 1, -3 and -72.

Write them down like this and drop the first coefficient below the horizontal line:

2 | 2 -3 1 -3 -72
|
------------------
2

Answer 9

Now multiply the first 2 with the number below the line (also 2) and put the result (4) in the next column and add:

2 | 2 -3 1 -3 -72
| 4
------------------
2 1

Answer 10

Multiply the first 2 with 1, put the result in the next column and add:

2 | 2 -3 1 -3 -72
| 4 2
------------------ +
2 1 3

Keep doing this until the end.

Answer 11

The remainder is -66 by the way, so i let you look if you get the same with your method Zehanz.

Answer 12

2 | 2 -3 1 -3 -72
| 4 2 6 6
------------------ +
2 1 3 3 ...

The number that will appear on the dots is the remainder!

Answer 13

So the remainder is: -66.

We now could write that division as:

\(\dfrac{2x^4-3x^3+x^2-3x-72}{x-2}=2x^3+x^2+3x+3-\dfrac{66}{x-2}\),

because the numbers we found below the horizontal line in the synthetic division are the coefficients of the quotient.

Answer 14

-66

Answer 15

Of course, we could just substitute x=2 in 2x4 – 3x3 + x2 – 3x – 72, to get -66, but doing the synth dev gives you something extra!