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--------------------------------------------------------------------- find the square root of x^4 -6(x^3) + 19(x^2) -30x + 20 first split into groups of 2 terms each from the last term x^4 + [ -6(x^3) + 19(x^2)] + [ -30x + 20] now first term is x^4 which is a perfect square x^4 = [x^2]^2 using long division method first step gives remainder 0 then the next group [ -6(x^3) + 19(x^2)] is taken down current quotient is [x^2] multiply it by 2 get 2[x^2] so the the next divisor is of the form 2[x^2] + ? for getting the unknown term ? divide the leading term in the group -6(x^3) with 2[x^2] to get -3x so the new divisor is 2[x^2] -3x introduce (-3x) in the quotient. multiply{2[x^2] -3x}*{-3x} to get -6[x^3] +9x subtract to get 10[x^2] bring down the next group to get 10[x^2] -30x +20 multiply the current quotient [x^2] - 3x with 2 to get 2[x^2] - 6x so new divisor is of the form 2[x^2] - 6x + ? divide the leading terms 10[x^2] with 2[x^2] to get 5 so new divisor is 2[x^2] - 6x +5 introduce +5 in the quotient multiply 2[x^2] - 6x +5 with 5 to get 10[x^2] - 30x + 25 subtract to get the remainder 0 and the square root | [x^2] -3x +5 | other questions and problems: * integral
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