x3 - 6x2 + 7x - 2 = x3 - x2 - 5x2 + 5x + 2x - 2
= x2(x - 1) - 5x(x - 1) + 2(x - 1)
= (x - 1)(x2 - 5x + 2)
= (x - 1){x - 0.5*[5 - sqrt(25 - 8)]}){x + 0.5*[5 - sqrt(25 - 8)]}
= (x - 1)(x - 0.4384)(x - 4.5616)
so that the roots are
x = 1
x = 0.4384
and
x = 4.5616
Since there is nothing following, none of them!
(2x - 9)(2x - 9)
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
To find the roots of the polynomial (x^2 + 5x + 9), we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}). Here, (a = 1), (b = 5), and (c = 9). The discriminant (b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 9 = 25 - 36 = -11), which is negative. This means the polynomial has no real roots, but two complex roots: (x = \frac{-5 \pm i\sqrt{11}}{2}).
The actual equation itself is the polynomial. There is no polynomial for it, and your question doesn't really make sense.
1
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
None, it involves the square root of a negative number so the roots are imaginary.
-2.5 + 1.6583123951777i-2.5 - 1.6583123951777i
A 7th degree polynomial.
You can find the roots with the quadratic equation (a = 1, b = 3, c = -5).
There are none because the discriminant of the given quadratic expression is less than zero.
Since there is nothing following, none of them!
(2x - 9)(2x - 9)
15j2(j + 2)
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
x2 + 12x + 32 = (x + 8)(x + 4)