Cubic equations

Here are the solutions for the cubic equations: 1) Solve x^3 - 2x^2 - 5x + 6 = 0 Step 1: Find possible rational roots using the Rational Root Theorem. The constant term is 6, and the leading coefficient is 1. Possible rational roots are the divisors of 6: ± 1, ± 2, ± 3, ± 6. Step 2: Test for a root using substitution or synthetic division. Let P(x) = x^3 - 2x^2 - 5x + 6. Test x=1: P(1) = (1)^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0 Since P(1)=0, x=1 is a root, and (x-1) is a factor. Step 3: Use synthetic division to find the depressed quadratic equation. Dividing x^3 - 2x^2 - 5x + 6 by (x-1): ` 1 | 1 -2 -5 6 | 1 -1 -6 ---------------- 1 -1 -6 0 ` The resulting quadratic equation is x^2 - x - 6 = 0. Step 4: Solve the quadratic equation. Factor the quadratic equation: (x-3)(x+2) = 0 This gives the roots x=3 and x=-2. The solutions for the equation x^3 - 2x^2 - 5x + 6 = 0 are: x=1, x=3, x=-2 2) Solve x^3 - 6x^2 + 11x - 6 = 0 Step 1: Find possible rational roots using the Rational Root Theorem. The constant term is -6, and the leading coefficient is 1. Possible rational roots are the divisors of -6: ± 1, ± 2, ± 3, ± 6. Step 2: Test for a root using substitution or synthetic division. Let P(x) = x^3 - 6x^2 + 11x - 6. Test x=1: P(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 Since P(1)=0, x=1 is a root, and (x-1) is a factor. Step 3: Use synthetic division to find the depressed quadratic equation. Dividing x^3 - 6x^2 + 11x - 6 by (x-1): ` 1 | 1 -6 11 -6 | 1 -5 6 ---------------- 1 -5 6 0 ` The resulting quadratic equation is x^2 - 5x + 6 = 0. Step 4: Solve the quadratic equation. Factor the quadratic equation: (x-2)(x-3) = 0 This gives the roots x=2 and x=3. The solutions for the equation x^3 - 6x^2 + 11x - 6 = 0 are: x=1, x=2, x=3 3 done, 2 left today. You're making progress.