70 likes | 261 Views
Aim: How do we solve higher degree equations?. Do Now: Factor the following. 1. x 3 – 4 x 2 – 5 x. 2. x 4 – 81. 3. x 3 – 2 x 2 + 3 x – 6. x 3 – 4 x 2 – 5 x = 0. This is a higher degree equation. Factor by GCF. x ( x 2 – 4 x – 5) = 0. x ( x – 5)( x + 1) = 0.
E N D
Aim: How do we solve higher degree equations? Do Now: Factor the following 1. x3 – 4x2 – 5x 2. x4 – 81 3. x3 – 2x2 + 3x – 6
x3 – 4x2 – 5x = 0 This is a higher degree equation • Factor by GCF x(x2 – 4x – 5) = 0 x(x – 5)(x + 1) = 0 Factor the trinomial Set each factor = 0 then solve x = 0, x – 5 = 0, x = 5, x + 1 = 0, x = -1
Solve for x: x3 + 7x2 + 10x = 0 x(x2+ 7x + 10) = 0 • Factor by GCF x(x+ 5)(x + 2) = 0 Factor the trinomial x = 0, x + 5 = 0, x = -5, x + 2 = 0, x = -2 Set each factor = 0 then solve
Solve for x: x3 – 3x2 – 4x + 12 = 0 Set it into 2 groups (x3 – 3x2) – (4x – 12) = 0 x2(x – 3) – 4(x – 3) = 0 Factor the GCF Factor the GCF again (x – 3)(x2 – 4) = 0 Factor completely (x – 3)(x – 2)(x + 2) = 0 x – 3 = 0 , x = 3 x – 2 = 0, x = 2 x + 2 = 0, x = -2 Set each factor = 0, then solve
Solve for x: x3 – 2x2– 9x+ 18 = 0 Set it into 2 groups (x3 – 2x2) – (9x– 18) = 0 x2(x – 2) – 9(x – 2) = 0 Factor the GCF Factor the GCF again (x – 2)(x2 – 9) = 0 Factor completely (x – 2)(x – 3)(x + 3) = 0 x – 2 = 0 , x = 2 x – 3 = 0, x = 3 x + 3 = 0, x = -3 Set each factor = 0, then solve
Solve for x: x4 – 5x2 + 4 = 0 Factor it as a trinomial with base = x2 (x2 – 4)(x2 – 1) = 0 (x2 – 4) = 0, (x2 – 1) = 0 Set each factor = 0, Factor each binomial then solve (x – 2)(x – 2) = 0 x = 2, – 2 (x– 1)(x + 1) = 0 x = 1, – 1
Solve the variable: 1. x3 – 8x2 – 9x = 0 2. x3 + 3x2– 4x– 12 = 0 3. 2x3 – 3x2 – 2x + 3 = 0 4. x4 – 13x2 + 36 = 0 5. x4 – 1 = 0