Solving higher degree equations Algebraically Lesson 2.6

1. Solving higher degree equations Algebraically Lesson 2.6 Solve: x3 + 5x2 – 4x – 20 = 0 (Hint: Factor by grouping) (x3 + 5x2) – (4x + 20) = 0 Group and factor out -1 from last two terms x2(x + 5) – 4(x + 5) = 0 (x + 5)(x2 – 4) = 0 continue factoring the difference of squares! (x + 5)(x+2)(x – 2) = 0 Ta da: x = - 5, - 2, & 2 Solve: 2x2 – x – 3 = 0 (Solve -- ?? would you go about this one?) It’s quadratic – try factoring  yipee!! (2x – 3)(x + 1) = 0 Dah  x = 3/2 & x = - 1 Now apply what you have learned to the next situation 

2. Now try to relate 2x4 – x2 – 3 = 0 (solve ??) (Is this in quadratic form? —(Hm? – there are only 3 terms and the middle terms power is half of the leading terms power, and the last term is a constant?) This sounds like quadratic to me    Soooo -- If we rewrite the equation like this -- 2(x2)2 – (x2) – 3 = 0 And let x2 = ?? Oh say ‘n’ then we get 2(n)2– (n) – 3 = 0 (Now factor in terms of n ) (2n – 3)(n + 1) = 0 n = 3/2, n = - 1 (Now re-replace n with x2 - yields) x2 = 3/2 x2 = -1 √x2 = √3/2, √x2 = √- 1 (now solve each for ‘x’)  x = +(√6)/4 , or x = +i I am sooo confused  ! Know you’re knot – you can handle this  Continue  Solve: x4 – 3x2 – 4 = 0 Dah  just like the last one – put in quadratic form (x2)2 – 3(x2) – 4 = 0 Replace x2 with ‘n’ n2 – 3n – 4 = 0 Now go for it!

3. (n – 4)(n + 1) = 0 n = 4 and n = -1 Re-replace ‘n’ with x2 - x2 = 4 and x2 = - 1 √x2 = √4 and √x2 = √- 1 x = + 2 & x = + I Solve: x3 + 6x2 – 4x – 24 = 0 ?? Any ideas?? Try factor by grouping and go for it!    (x3 + 6x2) - (4x + 6) = 0 x2(x + 6) – 4(x + 6) = 0 (x + 6)(x2 – 4) = 0 (x + 6)(x + 2)(x – 2) = 0 finally  x = - 6, - 2 & 2