Question

Помогите решить №1 Разделите уголком и по схеме Горнера многочлен: а) x^4 - 81 на х+3; x-3; x+1 №2 Решите уравнение а) (х + 1) (2х - 3) = 0; б) (Зх + 1) (х - 2) = 0; в) (х^2 - 1) (х + 3) = 0 Буду очень благодарна

Answer 1

1ое задание на фото не стал париться :D

2ое задание
а)уравнение тогда равно нулю, когда одна из скобок равна нулю

х+1=0______2х-3=0
х=-1_______2х=3
__________х=1.5
Корни уравнения: -1, 1.5

б)(3х+1)(х-2)=0 по такому же принципу :)
3х+1=0______х-2=0
3х=-1________х=2
х=-⅓

Корни уравнения: -⅓, 2

в) (Х^2-1)(х+3)=0

Х^2-1=0_____х+3=0
х^2=1 ______х=-3

х1=1 х2=-1

Корни уравнения : 1,-1,-3

Answer 2

Solving Polynomial Division Using Synthetic Division and the Rational Root Theorem

To divide the polynomial x^4 - 81 by x - 3, we can use synthetic division. The rational root theorem states that if a polynomial has a rational root, then it will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In this case, the constant term is 81, and the leading coefficient is 1. The factors of 81 are ±1, ±3, ±9, ±27, and ±81, and the factors of 1 are ±1. Therefore, the possible rational roots are ±1, ±3, ±9, and ±27. We can use synthetic division to test these roots.

The synthetic division process is as follows: 1. Write down the coefficients of the polynomial in order. 2. Bring down the first coefficient. 3. Multiply the root by the result from step 2 and write the answer below the next coefficient. 4. Repeat step 3 until all coefficients have been multiplied and added.

Let's perform the synthetic division to divide the polynomial x^4 - 81 by x - 3:

The coefficients of the polynomial are: 1, 0, 0, 0, -81 The root we are testing is: 3

Performing the synthetic division: ``` 3 | 1 0 0 0 -81 | 3 9 27 81 ----------------- 1 3 9 27 0 ``` The result of the synthetic division is: x^3 + 3x^2 + 9x + 27 with a remainder of 0.

Therefore, the division of x^4 - 81 by x - 3 yields the quotient x^3 + 3x^2 + 9x + 27.

Solving Equations Using Factoring and the Quadratic Formula

Now, let's solve the given equations:

a) (x + 1)(2x - 3) = 0 To solve this equation, we can set each factor equal to zero and solve for x: x + 1 = 0 => x = -1 2x - 3 = 0 => 2x = 3 => x = 3/2

So, the solutions to the equation are: x = -1 and x = 3/2.

b) (3x + 1)(x - 2) = 0 Setting each factor equal to zero and solving for x: 3x + 1 = 0 => 3x = -1 => x = -1/3 x - 2 = 0 => x = 2

The solutions to the equation are: x = -1/3 and x = 2.

c) (x^2 - 1)(x + 3) = 0 Factoring the quadratic equation x^2 - 1 gives us: (x - 1)(x + 1) Setting each factor equal to zero and solving for x: x - 1 = 0 => x = 1 x + 1 = 0 => x = -1

The solutions to the equation are: x = 1 and x = -1.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.