(2x^3+2x^2-16x+32)÷(x^2-3x+4) divide polynomial using long division

                    
                   
x^2-3x+4  /    2x^3+2x^2-16x+32

Note that x^2 divides into 2x^3 with a partial quotient of 2x.  Multiply x^2-3x+4 by this 2x and write your product under    2x^3+2x^2-16x+32:

                    
                  __2x 
x^2-3x+4  /    2x^3+2x^2-16x+32
                      2x^3 -6x^2 + 8x
                         subtract
                              8x^2 -24x + 32

Now divide x^2 into 8x^2:  result is 8.

                  __2x + 8
x^2-3x+4  /    2x^3+2x^2-16x+32
                      2x^3 -6x^2 + 8x
                         
                              8x^2 -24x + 32
                               8x^2 -24x + 32
                               subtract
                                                     0

So the desired quotient is 2x+8, or 2(x+4).

The full question isn’t shown. part of it is cut off

If jon had $4.00 then takes away $2.25, then he should have $1.75 left. $4.00 - $2.25 = $1.75 hope this ; )