Question

P(x) = х³ - 9х² + 27х - 27 = 0

Answer 1

Объяснение:

P(x) = х³ - 9х² + 27х - 27 = 0

х³ - 9х² + 27х - 27 = 0

х³ - 27 - 9х² + 27х = 0

(х³ - 27 )- (9х² - 27х) = 0

(х³ - 27 )- 9 x(х - 3) = 0

(x-3)( x^2 +3x +9) - 9x( x-3)= 0

(x-3)(x^2 +3x-9x+9)=0

(x-3 ) ( x^2 -6x +9)=0

x-3=0

x1=3

x^2 -6x +9=0

D= (-6)^2 -4*9=36-36=0

x= 6/2=3

(x-3)^2=0

x-3=0

x=3

ответ 3

Answer 2

To solve the equation P(x) = x³ - 9x² + 27x - 27 = 0, we can use various methods such as factoring, synthetic division, or numerical methods like Newton's method. Let's try to solve it using the rational root theorem and synthetic division.

The rational root theorem states that any rational root of a polynomial equation with integer coefficients must be of the form p/q, where p is a factor of the constant term (in this case, -27) and q is a factor of the leading coefficient (in this case, 1). Let's find the factors of -27 and 1.

Factors of -27: ±1, ±3, ±9, ±27 Factors of 1: ±1

Therefore, the possible rational roots are: ±1, ±3, ±9, ±27

Now, we can perform synthetic division to check if any of these possible roots are actual roots of the polynomial equation.

Let's start with the possible root x = 1:

1 | 1 -9 27 -27 | 1 -8 19

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     1   -8    19    -8

The remainder is -8, not zero. Therefore, x = 1 is not a root of the equation.

Now, let's try x = -1:

-1 | 1 -9 27 -27 | -1 10 -17

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     1  -10    37   -44

The remainder is -44, not zero. Therefore, x = -1 is not a root of the equation.

We can continue this process with the remaining possible roots, but in this case, none of them are actual roots of the equation. Therefore, the equation x³ - 9x² + 27x - 27 = 0 does not have any rational solutions.

To find the solutions, we may need to use numerical methods like the Newton-Raphson method or approximate them using graphing techniques or calculators.