If the roots of ${\mathrm{x}}^4 – 8{\mathrm{x}}^3 + 14{\mathrm{x}}^2 + 8\mathrm{x} – 15 = 0$ are in A.P. then the roots are

$\begin{array}{l}Leta-3d,a-d,a+d,a+3darerootsof{x}^4-8x^3+14x^2+8x-15=0\\ \therefore a-3d+a-d+a+d+a+3d=8\\ \Rightarrow 4a=8\Rightarrow a=2\\ \therefore (a-3d)(a-d)(a+d)(a+3d)=-15\\ \Rightarrow (a^2-9d^2)(a^2-d^2)=-15\\ \Rightarrow (4-9d^2)(4-d^2)=-15\\ \Rightarrow 16-4d^2-36d^2+9d^4=-15\\ \Rightarrow 9d^4-40d^2+31=0\\ \Rightarrow 9d^4-31d^2-9d^2+31=0\\ \Rightarrow d^2(9d^2-31)-1(9d^2-31)=0\\ \Rightarrow (d^2-1)(9d^2-31)=0\\ \Rightarrow d^2=1,d^2=\frac{31}{9}\\ \therefore d=1\\ Therootsare2-3,2-1,2+1,2+3\\ =-1,1,3,5\end{array}$