Find the square root of \( 4 x^{4}-12 x^{3}+29 x^{2}-30 x+9 \)

We wish to find an expression for \[ \sqrt{4x^4-12x^3+29x^2-30x+9}\,. \] A common method is to try to write the radicand as a perfect square minus a constant so that we can use the difference‐of‐squares factorization. 1. Notice that if we try to express \[ 4x^4-12x^3+29x^2-30x+9 = (2x^2-3x+A)^2 + \text{(extra terms)}, \] then we must expand \[ (2x^2-3x+A)^2 = 4x^4 - 12x^3 + (9+4A)x^2 - 6Ax + A^2\,. \] We compare coefficients with the given polynomial. 2. To match the \(x^2\) term we set \[ 9+4A = 29\quad\Longrightarrow\quad 4A=20\quad\Longrightarrow\quad A=5\,. \] Now the expansion gives: \[ (2x^2-3x+5)^2 = 4x^4-12x^3+ (9+20)x^2- 30x+25 = 4x^4-12x^3+29x^2-30x+25\,. \] Notice that this differs from the given polynomial in the constant term: \[ (2x^2-3x+5)^2 \text{ has constant } 25,\quad \text{but we need }9\,. \] 3. In fact, \[ (2x^2-3x+5)^2-16 = 4x^4-12x^3+29x^2-30x+25-16 = 4x^4-12x^3+29x^2-30x+9\,. \] Therefore, we can write \[ 4x^4-12x^3+29x^2-30x+9=(2x^2-3x+5)^2-16\,. \] 4. Recognize the structure as a difference of two squares. Since \[ 16=(4)^2, \] we have \[ (2x^2-3x+5)^2-4^2 = \bigl[(2x^2-3x+5)-4\bigr]\bigl[(2x^2-3x+5)+4\bigr]\,. \] That is, \[ 4x^4-12x^3+29x^2-30x+9 = (2x^2-3x+1)(2x^2-3x+9)\,. \] 5. Thus one acceptable answer is to write the square root as \[ \sqrt{4x^4-12x^3+29x^2-30x+9} = \sqrt{(2x^2-3x+1)(2x^2-3x+9)}\,. \] Any answer equivalent to the above is correct.