Solve Quadratic Equation (x³-4x²+5x-2):(x-1)= x²- 2x-8 | Need Help?

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The discussion focuses on solving the polynomial division of (x³ - 4x² + 5x - 2) by (x - 1) and clarifies the process of polynomial long division. It emphasizes that the division of x³ by x results in x², not 3x², and explains the step-by-step approach to finding the quotient. The analogy of dividing numbers is used to illustrate the division process, highlighting that the rules of algebra apply similarly to polynomial terms. The conversation encourages practicing the same method with the given polynomial to understand the division better. Mastering this technique is essential for accurately solving polynomial equations.
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here is my task: (x³-4x²+5x-2):(x-1)=
We did such task in school but I didn't got it ..

Here is the task that we did in school
(x³-3x²-6x+8) :(x-1)= x²- 2x-8 but I don`t get it why x² is the result and why 3x² ??
Can someone help me? Thank you in advance
 
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Regular polynomial division is easier than standard long division with pure constants. Your approach should be exactly the same.

Ask this: What is x^3 divided by x ? It is certainly not three times anything. You have no reason for coming to 3*x^2.
 
To expand on symbolipoint's response, suppose you were asked to divide 3524232 by 83. You would start by noting that 8 divides into 35 for times so the first digit of the quotient is (probably) 4. you need the "probably" because the digit after the 8 might cause a problem. But in this case 4*83= 329 and 352- 329= 23, less than 83. So you subtract those and your problem becomes 234232 divided by 81. 8 will divide into 23 twice (3*8= 24 so 3 times doesn't quite fit). 2*83= 163 and 234- 163= 71. You know that the first two digits of the quotient are 42.. and you continue.

x3- 3x2- 6x- 8 divided by x-1 is much the same thing- the "x" just represents a number so all the rules of algebra apply. x divides into x3 x3/x= x2[/sup times so we try x2 as the divisor: x2 times x-1 is x3- x2 and x3- 3x2 minus x3- x2 is -3x2- (-x[sup[2)= -2x2. Of course, the x3 terms cancel- that was the whole point of choosing the divisor to be x2. We now have left -2x2- 6x+ 8 to be divided by x- 1.

x2 will divide into -2x2 -2x times, so we try -2x as the next term in the quotient: -2x times x- 1 is -2x2+ 2x and -2x2- 6x minus -2x2+ 2x is just -6x-(2x)= -8x. We have left -8x+ 8 to be divided by x- 1.

Okay, -8x divided by x is, of course, -8 and -8 times x-1 is -8+ 8, exactly what we had left: x-1 divides into -8x+ 8 exactly -8 times. Putting all of that together, x-1 divides into x3- 3x2+ 6x- 8 exactly x2-2x -8 times.

Now, you try exactly the same thing with x3- 4x2+ 5x- 2 divided by x-1. Warning: even when you are dividing numbers, divisions don't always come out even: 33 divided by 4 is 8 with a remainder of 1 (or 8+ 1/4).
 
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