Question 723736: 2x^4-3x^3+4x^2-9/
2x-3
Divide. Check you answer if there is a remainder put it over the the divisor! Thanks
^ means its a exponent. Answer by DrBeeee(684) (Show Source):
You can put this solution on YOUR website! Given:
(1) (2x^4-3x^3 + 4x^2-9)/(2x-3)
Let's separate the numerator of (1) into the sum of two binomials and get
(2) [(2x^4-3x^3) + (4x^2-9)]/(2x-3)
Now factor the two numerator binomials and get
(3) [(x^3)*(2x-3) + (2x-3)*(2x+3)]/(2x-3)
Now carry out the division of each numerator term by the denominator (2x-3) and get
(4) (x^3)*(2x-3)/(2x-3) + (2x-3)*(2x+3)/(2x-3) or after cancellation we get
(5) x^3 + 2x + 3
Let's check this.
Is ((x^3+2x+3)*(2x-3) = 2x^4-3x^3+4x^2-9)?
Is (2x^4+4x^2+6x-3x^3-6x-9 = 2x^4-3x^3+4x^2-9)?
Is (2x^4-3x^3+4x^2-9 = 2x^4-3x^3+4x^2-9)? Yes
Answer: The given expression simplifies to x^3 + 2x + 3.
PS there is no remainder.
(