Question

Divide:(10x²-3x+4)÷(2x-5)

This is a review homework for Algebra II the question is about Algebra I. i would like an explanation please.
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Answer 1

`(10x^2-3x+4)/(2x-5)=5x+11+(59)/(2x-5)`

Quotient: 5x+11

Remainder: 59

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:

`5x+11+(59)/(2x-5)`.

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:

`(10x^2-3x+4)/(2x-5)`

`=(10x^2-3x+4)/(2(x-(5)/(2)))=(1)/(2) \cdot (10x^2-3x+4)/(x-(5)/(2))=(1)/(2)(10x+22+(59)/(x-(5)/(2)))`

Distribute the 1/2 back:

`(10x^2-3x+4)/(2x-5)=(10x+22)/(2)+(59)/(2(x-(5)/(2)))`

`(10x^2-3x+4)/(2x-5)=5x+11+(59)/(2x-5)`