The polynomial $x^4+a{x}^2+b$ is exactly divisible by $x^2-4x+3$. Then $b$ equals:

Since $x^2-4x+3=(x-1)(x-3)$, both 1 and 3 must be zeros: $1+a+b=0$ and $81+9a+b=0$. Subtracting, $80+8a=0$, so $a=-10$ and then $b=9$. (−10 is the value of a, not of $b$; the sign-flipped options come from moving a term across the equals sign carelessly.) The polynomial is $x^4-10x^2+9=(x^2-1)(x^2-9)=(x-1)(x+1)(x-3)(x+3)$, which visibly contains $(x-1)(x-3)$.