$$(x+1)(x+3)(x-2)(x-6)+96$$
$\left(x-6\right)\left(x-2\right)\left(x+1\right)\left(x+3\right)+96$
$$\left(x^{2}+3x+x+3\right)\left(x-2\right)\left(x-6\right)+96$$
$$\left(x^{2}+4x+3\right)\left(x-2\right)\left(x-6\right)+96$$
$$\left(x^{3}-2x^{2}+4x^{2}-8x+3x-6\right)\left(x-6\right)+96$$
$$\left(x^{3}+2x^{2}-8x+3x-6\right)\left(x-6\right)+96$$
$$\left(x^{3}+2x^{2}-5x-6\right)\left(x-6\right)+96$$
$$x^{4}-6x^{3}+2x^{3}-12x^{2}-5x^{2}+30x-6x+36+96$$
$$x^{4}-4x^{3}-12x^{2}-5x^{2}+30x-6x+36+96$$
$$x^{4}-4x^{3}-17x^{2}+30x-6x+36+96$$
$$x^{4}-4x^{3}-17x^{2}+24x+36+96$$
$$x^{4}-4x^{3}-17x^{2}+24x+132$$
Show Solution
Hide Solution
$x^{4}-4x^{3}-17x^{2}+24x+132$