x^4-4x^2+3=0

|2x-5|=11

For the first equation don’t let the cube trip you up. simply factor out 5x because 5 goes into all three numbers evenly as does the x. so now your equation reads 5x(x²+6x+9)=0. now factor x²+6x+9 like you normally would. now you should have **3 possible roots. **5x=0, x+3=0, and x+3=0. once you solve for x you should have x=0 and x=-3.

for the second one its a little trickier. we cant factor out the way we did in number one so you try to get all the x’s to one side. ⇒ x^4-4x²=-3. now you can factor x² out to get x²(x²-4)=-3. now solve for x! x²=-3 and x²-4=-3. you get x=√-3, x=1 and x=-1

for the last one your going to solve the original version of the problem (2x-5=11) and the negated version of the problem. (-2x+5=11) all you’re doing is solving for x. you should get x=-3 and x=8

\( 5x^3+30x^2+45x=0 \\ x^3+6x^2+9x=0\\ x(x^2+6x+9)=0\\ x(x+3)^2=0\\ x=0 \vee x=-3\\\\ x^4-4x^2+3=0\\ x^4-x^2-3x^2+3=0\\ x^2(x^2-1)-3(x^2-1)=0\\ (x^2-3)(x^2-1)=0\\ (x^2-3)(x-1)(x+1)=0\\ x=-\sqrt3 \vee x=\sqrt 3 \vee x=1 \vee x=-1 \)

\( |2x-5|=11\\ 2x-5=11 \vee 2x-5=-11\ 2x=16 \vee 2x=-6\\ x=8 \vee x=-3 \)