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 .

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:

.

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:

Distribute the 1/2 back:

answer: the answer is 5/22

step-by-step explanation: if you subtract 17 from 22 the answer is 5 therefore the remainder of students are female

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x