2. The polynomial \( 6 y^{4}-19 y^{3}-23 y^{2}+10 y+8 \) on factorization gives: (a) \( (y+1)(y-4)(3 y+2)(2 y+1) \) (b) \( (y+1)(y-4)(3 y-2)(2 y-1) \) (c) \( (y+1)(y-4)(3 y-2)(2 y+1) \) (d) \( (y+1)(y-4)(3 y+2)(2 y-1) \)

To factorize the polynomial \(6y^{4} - 19y^{3} - 23y^{2} + 10y + 8\), we need to find the roots of the polynomial and then express the polynomial as a product of linear factors. Let's find the roots of the polynomial by setting it equal to zero and solving for \(y\): \[6y^{4} - 19y^{3} - 23y^{2} + 10y + 8 = 0\] Now, we will factorize the polynomial using the roots we find. Let's solve for the roots first. Solve the equation by following steps: - step0: Solve for \(y\): \(6y^{4}-19y^{3}-23y^{2}+10y+8=0\) - step1: Factor the expression: \(\left(y-4\right)\left(y+1\right)\left(2y+1\right)\left(3y-2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&y-4=0\\&y+1=0\\&2y+1=0\\&3y-2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&y=4\\&y=-1\\&y=-\frac{1}{2}\\&y=\frac{2}{3}\end{align}\) - step4: Rewrite: \(y_{1}=-1,y_{2}=-\frac{1}{2},y_{3}=\frac{2}{3},y_{4}=4\) The roots of the polynomial \(6y^{4} - 19y^{3} - 23y^{2} + 10y + 8\) are \(y = -1\), \(y = -\frac{1}{2}\), \(y = \frac{2}{3}\), and \(y = 4\). Now, we will express the polynomial as a product of linear factors using these roots. The factors will be in the form \((y - r)\), where \(r\) is a root of the polynomial. The factors will be: \((y + 1)\), \((y + \frac{1}{2})\), \((y - \frac{2}{3})\), and \((y - 4)\). Therefore, the factorization of the polynomial \(6y^{4} - 19y^{3} - 23y^{2} + 10y + 8\) is \((y + 1)(y + \frac{1}{2})(y - \frac{2}{3})(y - 4)\). Comparing this with the given options, the correct factorization is \((y + 1)(y - 4)(3y + 2)(2y + 1)\), which corresponds to option (a).