2x2+10x+12/(x+3)
We use long division just like we would normally.
Let me give you an example.
ex:
Divide 7 into 23
---------
7 | 234
We say ok, how many times does 7 go into 2?
0 times, it doesn't work. So you put a 0 above the 2.
Now we try 7 into 23. It can go 3 times, so we put a 3 above the 3 (in 234) and subtract 21. (3 *7) and then carry down the 4.
...03
--------
7|234
-..21
--------
...024
how many times will 7 go into 24? 3 again.
...033
-------
7|234
-..21
-------
...024
-....21
--------
.......3
We're left with 3 as the remainder.
Our answer to 234 divided by 7 is 33 remainder of 3, or 33 and 3/7.
----
This method works similarly with variables.
2x2+10x+12/(x+3)
----------------------
x+3|2x2+10x+12
We'll start the same way. How many times will X go into 2x2
Or, what times X gives 2x2 x * 2x = 2x2
So we write a 2x above the 2x2, just like we wrote the 3 in the example above.
.......2x
----------------------
x+3|2x2+10x+12
Now we multiply x+3 by 2x to figure out what to subtract.
2x(x+3)= 2x2 +6x
Notice we ended up with a 2x2? This is what we wanted to subtract! Something to note, when you do your subtraction, you're subtracting the entire expression 2x2 +6x.
So you can write -2x2 - 6x.
.......2x
----------------------
x+3|2x2+10x+12
......-2x2 - 6x
-----------------
...............4x + 12
Make sure you carry down the next term, the +12. Just like we carried down the 4 in the example above.
Now, how many times will X go into 4x? 4 times. So we write a 4 next to the 2x.
and then multiply 4(x+3). then subtract.
.......2x + 4
----------------------
x+3|2x2+10x+12
......-2x2 - 6x
-----------------
...............4x + 12
..............-4x - 12
---------------------
........................0
In this scenario we get a remainder of 0.
This means that x+3 divides evenly into 2x2+10x+12.
In fact, it can divide into it 2x +4 times.
To check this, multiply (x+3)(2x+4) use FOIL.
(x+3)(2x+4) = 2x2+10x+12 (check)
Side note: If you did get a remainder, like in the 1st example. Let's say the remainder was 1.
You take the remainder, 1 and put it over the divisor x +3.
so your answer would be 2x+4 +(1/x+3)
-9(x + 8)
36 36 is 6 squared. 6 times 6. 36 divided by 2 is 18 36 divided by 3 is 12
(1/2, 71 and 3/4)or(0.5, 71.75)
Yes, it can. ANY equation of this form can be factored. The factors may not be integers or even real numbers, though.
Suspended account squared
P squared = P*P. When divided by P, the equation becomes (P*P/P, and the answer is "P".
x3 -3x2 -x - 1 divided by x+2 equals x2-5x+9 remainder -19 It's difficult to show how to work it out on this computer but division with algebra has a lot in common with doing long division with integers.
no
Power (Joules) = the square root of the voltage squared divided by the resistance
The quadric equation is: negative b plus or minus the square root of b squared minus 4ac all over(divided by) 2a
It's a simple problem in long division and the answer is 2t + 1
You should square root both the top and bottom of the equation, which would give you (√2x)/x.
The equation for this problem is pi squared •2 divided by mass
a squared or a2
A quadratic equation.
Yes
(x3-3x2+6x+5)/(x-2) = x2-x+4 remainder 13 or as x2-x+4+13/(x-2) It's difficult to show the workings out on this computer but division in algebra has a lot in common when doing long division with integers.