Webx3=125 Three solutions were found : x = (-5-√-75)/2= (-5-5i√ 3 )/2= -2.5000-4.3301i x = (-5+√-75)/2= (-5+5i√ 3 )/2= -2.5000+4.3301i x = 5 Rearrange: Rearrange the equation by … WebPrecalculus Find the Roots (Zeros) f (x)=x^3-125 f (x) = x3 − 125 f ( x) = x 3 - 125 Set x3 −125 x 3 - 125 equal to 0 0. x3 − 125 = 0 x 3 - 125 = 0 Solve for x x. Tap for more steps... x = 5,− 5−5i√3 2,− 5+ 5i√3 2 x = 5, - 5 - 5 i 3 2, - 5 + 5 i 3 2
Solve by Factoring x^3-x=0 Mathway
WebNote: The quadratic portion of each cube formula does not factor, so don't waste time attempting to factor it. Yes, a 2 − 2ab + b 2 and a 2 + 2ab + b 2 factor, but that's because of the 2 's on their middle terms. ... = −1(x 3 + 125) Aha! Now what's inside the parentheses is a sum of cubes, which I can factor. WebПорядок Действий Множители и Простые Числа Дроби Длинная aрифметика Десятичные Экспоненты и Радикалы Соотношения и Пропорции Модуль Среднее Значение, Медиана и Режим Арифметика Научных Обозначений family and friends 3 unit 12
factorizar x^3+125
WebAlgebra Solve for x x^3=125 x3 = 125 x 3 = 125 Subtract 125 125 from both sides of the equation. x3 − 125 = 0 x 3 - 125 = 0 Factor the left side of the equation. Tap for more steps... (x−5)(x2 +5x+25) = 0 ( x - 5) ( x 2 + 5 x + 25) = 0 Webx^3- (125)=0 Step by step solution : Step 1 : Trying to factor as a Difference of Cubes: 1.1 Factoring: x3-125 Theory : A difference of two perfect cubes, a3 - b3 can be factored into (a-b) • (a2 +ab +b2) Proof : (a-b)• (a2+ab+b2) = a3+a2b+ab2-ba2-b2a-b3 = a3+ (a2b-ba2)+ (ab2-b2a)-b3 = a3+0+0-b3 = a3-b3 Check : 125 is the cube of 5 WebSince both terms are perfect squares, factor using the difference of squares formula, where and . Step 4. Simplify. Tap for more steps... Step 4.1. Rewrite as . Step 4.2. Factor. Tap for more steps... Step 4.2.1. Since both terms are perfect squares, factor using the difference of squares formula, where and . cook a flat iron steak