Solve the equation a 0=x^3-4x^2-21x b 2x^4-6x^3=12x^2-36x

1. Factor the cubic equation (a)<br /> $0 = x^3 - 4x^2 - 21x$; factor out $x$: $x(x^2 - 4x - 21) = 0$<br />2. Solve for $x$ in (a)<br /> $x = 0$ or $x^2 - 4x - 21 = 0$; solve quadratic using **$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$** with $a=1$, $b=-4$, $c=-21$<br /> Discriminant: $(-4)^2 - 4(1)(-21) = 16 + 84 = 100$<br /> Roots: $x = \frac{4 \pm 10}{2} \implies x = 7,\, x = -3$<br /><br />3. Rearrange and factor quartic equation (b)<br /> $2x^4 - 6x^3 - 12x^2 + 36x = 0$; factor out $2x$: $2x(x^3 - 3x^2 - 6x + 18) = 0$<br />4. Find rational roots of cubic (b)<br /> Try $x=2$: $8 - 12 - 12 + 18 = 2$; $x=-3$: $-27 - 27 + 18 + 18 = -18$; $x=3$: $27 - 27 - 18 + 18 = 0$; so $x=3$ is a root.<br />5. Factor cubic by synthetic division (b)<br /> Divide $x^3 - 3x^2 - 6x + 18$ by $(x-3)$: result is $x^2 - 6$<br />6. Solve remaining factors (b)<br /> $x^2 - 6 = 0 \implies x = \pm\sqrt{6}$; also $x=0$ from earlier factor.