Ahh, this brings back the ptsd of doing an inverse laplace transform 😊

2:09 That's the way I was taught but it's still pretty daunting... How about something like this as a simplification method? (11x²-26x+3)/(2x+1)(x+2)² =(11(x+2)²-70x-41)/(2x+1)(x+2)² =11/(2x+1)-(35(2x+1)+6)/(2x+1)(x+2)² =11/(2x+1)-35/(x+2)²-6/(2x+1)(x+2)² ... Just keep compensating at each step. I see that this problem is different from yours as I put in a +2 instead of -2... So, try again with your numbers: (11x²-26x+3)/(2x+1)(x-2)² =(11(x-2)²+18x-41)/(2x+1)(x-2)² =11/(2x+1)+(9(2x+1)-50)/(2x+1)(x-2)² =11/(2x+1)+9/(x-2)² -50/(2x+1)(x-2)² Going back to your problem... You have the system that can be easily reduced, From this: 11 | 1 0 2 -26 |-4 2 -3 3 | 4 1 -2 To this: 11 | 1 0 2 14 | 5 1 0 -23 | 0 3 -5 To this: 11 | 1 0 2 -41 | 0 1 -10 100| 0 0 25 => C=100/25=4 => B=-41+10(4)= -1 => A= 11-2(4)=3 Let's check that: Let y=(2x+1)(x-2) => 3/(2x+1)-1/(x-2)²+4/(x-2) =((3(x-2)²-(2x+1))+4y)/y(x-2) = (11x² -26x +3)/y(x-2) ✅😊 What is "x"? So, we apparently have: 11/(2x+1)+9/(x-2)² -50/y(x-2) =3/(2x+1)-1/(x-2)²+4/(x-2) => 8/(2x+1) +10/(x-2)² -4/(x-2)= 50/y(x-2) Therefore: (11x² -26x +3)/y(x-2) =(11x² -26x +3)(8/(2x+1) +10/(x-2)² -4/(x-2))/50 = 3/(2x+1)-1/(x-2)²+4/(x-2) So, if z=11x² -26x +3, then we have the system: (z(8/50)-3)/(2x+1)+(z(10/50)+1)/(x-2)² -(z(4/50)+4)/(x-2) =0 Multiply all by y(x-2): (z(8/50)-3)(x-2)²+(z(10/50)+1)(2x+1) -(z(4/50)+4)(x-2)(2x+1)=0 Expand & rearrange: z[(8/50)(x-2)²+(10/50)(2x+1)-(4/50)(x-2)(2x+1)]=3(x-2)²-(2x+1)+(4)(x-2)(2x+1) z[(8/50)(x²-4x²+4)+(2(4/50)+10/50)(2x+1)-(4/50)x(2x+1)]=11x²-18x+3 z[(-32/50)x²+(32/50)x+50/50]=11x²-18x+3 z[(-16)x²+(16)x+25]=25(11x²-18x+3) z[(-16)x²+(16)x]=25((11x²+3)-z)-18(25)x 16z[ -x+1]x=25(26-18)x z[ -x+1]=25(8/16) 11x² -26x +3= 25/2(1-x) (11x -26)(2-2x) -6= 19/x 22x³ -74x²+58x+19=0 => x≈-0.245251, y≈-1.14395, z≈10.0381 & the original quotient should be: z/y(x-2)≈3.90823

For the first problem, C = 4, not 2

bruh is this not the normal way ppl do it

@TheUnqualifiedTutor

2 ай бұрын

nah, my teacher did some weird s*** where you let (x - a) = 0 and substitute in x. Some teachers do it this way though