Question

2x⁴-x³-14x² + 19x - 6=0​

Answer 1

Ответ:

Объяснение:

Отгадываем корни наугад, а после раскладываем на множители.

Answer 2

To solve the equation 2x⁴ - x³ - 14x² + 19x - 6 = 0, you can use various methods such as factoring, the Rational Root Theorem, or numerical methods. In this case, I will use the Rational Root Theorem to find the rational solutions (if they exist) and then use synthetic division to test those potential roots.

The Rational Root Theorem states that any rational root of the polynomial is of the form p/q, where p is a factor of the constant term (-6) and q is a factor of the leading coefficient (2). So, the possible rational roots are:

±1, ±2, ±3, ±6

Now, we will use synthetic division to test these potential roots to find if they are roots of the equation:

For x = 1: 2 * 1⁴ - 1³ - 14 * 1² + 19 * 1 - 6 = 2 - 1 - 14 + 19 - 6 = 0

So, x = 1 is a root.

Now, we can use polynomial long division or synthetic division again to factor the polynomial:

Dividing the polynomial 2x⁴ - x³ - 14x² + 19x - 6 by (x - 1):

Copy code
 2x³ + x² - 15x + 6

(x - 1) | 2x⁴ - x³ - 14x² + 19x - 6

Now, let's try to factor the cubic polynomial 2x³ + x² - 15x + 6. You can use various methods such as factoring by grouping or the Rational Root Theorem again. In this case, you can find that (x - 2) is a root:

2(2)³ + 2² - 15(2) + 6 = 16 + 4 - 30 + 6 = -4 + 6 = 2

So, x = 2 is a root.

Dividing 2x³ + x² - 15x + 6 by (x - 2):

Copy code
  2x² + 5x - 3

(x - 2) | 2x³ + x² - 15x + 6

Now, let's try to factor the quadratic polynomial 2x² + 5x - 3. You can use factoring or the quadratic formula to find the remaining roots. Factoring the quadratic, we get:

(2x - 1)(x + 3)

Setting each factor equal to zero:

2x - 1 = 0 2x = 1 x = 1/2

x + 3 = 0 x = -3

So, the roots of the original equation 2x⁴ - x³ - 14x² + 19x - 6 = 0 are x = 1, x = 2, x = 1/2, and x = -3.