Answer:
[tex]\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}[/tex]
Quotient: 5x+11
Remainder: 59
Step-by-step explanation:
I'm going to do long division.
The bottom goes on the outside and the top goes in the inside.
Setup:
---------------------------------
2x-5 | 10x^2 -3x +4
Starting the problem from the setup:
5x +11 (I put 5x on top because 5x(2x)=10x^2)
--------------------------------- (We are going to distribute 5x to the divisor)
2x-5 | 10x^2 -3x +4
-(10x^2 -25x) (We are now going to subtract to see what's left.)
-----------------------------------
22x +4 (I know 2x goes into 22x, 11 times.)
( I have put +11 on top as a result.)
-(22x -55) (I distribute 11 to the divisor.)
-----------------------
59 (We are done since the divisor is higher degree.)
The quotient is 5x+11.
The remainder is 59.
The result of the division is equal to:
[tex]5x+11+\frac{59}{2x-5}[/tex].
We can actually use synthetic division as well since the denominator is linear.
Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:
2x-5=0
Add 5 on both:
2x=5
Divide both sides by 2:
x=5/2
Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:
So 5/2 will go on the outside:
5/2 | 10 -3 4
| 25 55
------------------------------
10 22 59
So we have:
[tex]\frac{10x^2-3x+4}{2x-5}[/tex]
[tex]=\frac{10x^2-3x+4}{2(x-\frac{5}{2})}=\frac{1}{2} \cdot \frac{10x^2-3x+4}{x-\frac{5}{2}}=\frac{1}{2}(10x+22+\frac{59}{x-\frac{5}{2}})[/tex]
Distribute the 1/2 back:
[tex]\frac{10x^2-3x+4}{2x-5}=\frac{10x+22}{2}+\frac{59}{2(x-\frac{5}{2})}[/tex]
[tex]\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}[/tex]
