Question 1011718
you can solve this by synthetic division or by long division.
see the following worksheet:


first workssheet divides using long division and using synthetic division.


the remainder is 4c^2 + 6c + 1
since the remainder is -1, you set 4c^2 + 6x + 1 equal to -1 and then solve for c.


you add 1 to both sides of the equation to get 4c^2 + 6x + 2 = 0


<img src = "http://theo.x10hosting.com/2015/122901.jpg" alt="$$$" </>


then you factor.


the second worksheet shows you the result of factoring 4c^2 + 6x + 2 = 0


the factored equation is (4c+2) * (c+1) = 0.


solve for c to get c = -1/2 or c = -1


this means that 2c = -1 or 2c = -2


those are the roots of the quadratic equation.


when 2c = -1, x + 2c = x - 1.


when 2c = -2, x + 2c = x - 2.


your divisors of x^2 - 3x + 1 are (x-1) and (x-2).


<img src = "http://theo.x10hosting.com/2015/122902.jpg" alt="$$$" </>


the next worksheet shows you long division by x-1 and by x-2.


<img src = "http://theo.x10hosting.com/2015/122903.jpg" alt="$$$" </>


the next worksheet shows you synthetic division by x-1 and by x-2.


<img src = "http://theo.x10hosting.com/2015/122904.jpg" alt="$$$" </>


if you need a lesson on how to do long division or how to do synthetic division or how to factor a quadratic, see the following refences.


<a href = "http://www.purplemath.com/modules/factquad.htm" target = "_blank">http://www.purplemath.com/modules/factquad.htm</a>


<a href = "http://www.purplemath.com/modules/polydiv2.htm" target = "_blank">http://www.purplemath.com/modules/polydiv2.htm</a>


<a href = "http://www.purplemath.com/modules/synthdiv.htm" target = "_blank">http://www.purplemath.com/modules/synthdiv.htm</a>