How can we solve this equation? $x^4-8x^3+24x^2-32x+16=0.$ $\begingroup$ Since there are a lot of powers of two appearing in the polynomial equation, try $x=2$ and find that it is a solution.
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14 сент. 2023 г. ... If x - 2 is a factor, what is the value of u? Explain how you know. 1. See answer.
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www.mathway.comTherefore, the value of the constant 'k' in the factored form is 2. The correct answer is an option (C). Learn more about the factorization of polynomial here Consider these functions: f(x) = 3x3 + 8x − 2 k(x) = 4x What is the value of k(f(x))?
brainly.comPolynomial Roots Calculator : ... In this case, the Leading Coefficient is 1 and the Trailing Constant is 16. The factor(s) are: of the Leading Coefficient : ... Therefore, the value of the constant 'k' in the factored form is 2. The correct answer is an option (C). Learn more about the factorization of polynomial here Consider these functions: f(x) = 3x3 + 8x − 2 k(x) = 4x What is the value of k(f(x))?
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x в степени 4+8*x в степени 3+16*x в степени 2. 4x3+24x2+32x=0 Решаем это уравнение Корни этого уравнения.
mrexam.ru24 апр. 2023 г. ... You can then use completing square method to solve the equation. x4−8x3+24x2−32x+16−16−14=0 ... The discriminant of the quadratic factor is 16 ... x в степени 4+8*x в степени 3+16*x в степени 2. 4x3+24x2+32x=0 Решаем это уравнение Корни этого уравнения.
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x^4-8*x^3+24*x^2-32*x+16=0 Главная / x^4-8*x^3+24*x^2-32*x+16=0
uchimatchast.ru22 февр. 2021 г. ... x = -2 + (3)^1/2 i. x + 2 = i (3)^1/2. (x + 2)^4 = 9. Expanding the L.H.S gives,. x^4 + 8x^3 + 24x^2 + 32x + 16 = 9. Thus the answer is 9. x^4-8*x^3+24*x^2-32*x+16=0 Главная / x^4-8*x^3+24*x^2-32*x+16=0
www.quora.comHow do you factor #8x^6-32x^5+4x^4#? · How do you factor #5x^3-20x#? · How do you factor #-x^2-9x+22#?. x^4-8*x^3+24*x^2-32*x+16=0 Главная / x^4-8*x^3+24*x^2-32*x+16=0
socratic.orgSolve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, ... x^4-8*x^3+24*x^2-32*x+16=0 Главная / x^4-8*x^3+24*x^2-32*x+16=0
mathsolver.microsoft.comx4 – 8x3 + 24x2 – 32x + 16 = (x – 2)(x – 2)(x – 2)(x – 2). x – 2 is a factor four times. The root 2 has a multiplicity of 4. x4 – 8x3 + 24x2 – 32x + 16 = 0. x^4-8*x^3+24*x^2-32*x+16=0 Главная / x^4-8*x^3+24*x^2-32*x+16=0
www.slps.org30 мар. 2014 г. ... You could factorise it, in the manner of (x−2)4=0. I saw those factors immediately. x^4-8*x^3+24*x^2-32*x+16=0 Главная / x^4-8*x^3+24*x^2-32*x+16=0
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Помогите пожалуйста с выполнением задания. 2-10 шифратор в коде 3321. с помощью производной нужно найти наименьшее и наибольшее значения функции на промежутке [-3;0] с помощью производной с решением пожалуйста ;) у’=4х³+8*3х²+24*2х+32=4х³+24х²+48х+32=4(.
otvet.mail.ruРассмотрим вариант решения задания из учебника Мордкович, Семенов 10 класс, Мнемозина a) у = х4 + 8х3 + 24x2 + 32x + 21, [-3; 0] г) у = 0,25x4 - 7/3х3 + 3,5, [-1; 2].
reshak.ruDetailed step by step solution for x^4-8x^3+24x^2-32x+16=0. Рассмотрим вариант решения задания из учебника Мордкович, Семенов 10 класс, Мнемозина a) у = х4 + 8х3 + 24x2 + 32x + 21, [-3; 0] г) у = 0,25x4 - 7/3х3 + 3,5, [-1; 2].
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