Solve f(x)=x^4−4x^3−9x^2+26x−30.

Roots for an algebraic equation are those values for a variable x, which satisfies a given function f(x). A root can also be called the zero of the algebraic equation. There can be either real roots or imaginary roots or complex roots(real part + imaginary part).

Let's solve the given equation: f(x) = x⁴ - 4x³ - 9x² + 26x - 30

We have to find a value of x for which the function f(x) = 0.

Applying hit and trail method. Checking for x = ±1, ±2, ±3, ±4, ±5.... etc.

We find that for x = -3, f(x) = 0.

so x = -3 is a root for the polynomial f(x) = x⁴ - 4x³ - 9x² + 26x - 30.

=> (x + 3 = 0) is a root for the polynomial f(x) = x⁴ - 4x³ - 9x² + 26x - 30.

Now factorizing f(x) = x⁴ - 4x³ - 9x² + 26x - 30, taking (x + 3) as common factor , we get:

f(x) = x⁴ + 3x³ - 7x³ - 21x² + 12x² + 36x - 10x - 20

f(x) = x³(x + 3) - 7x²(x + 3) + 12x(x + 3) - 10

f(x) = (x + 3)(x³ - 7x² + 12x - 10)

Now, we need to factorize this part (x³ - 7x² + 12x - 10).

Let us consider it to be g(x) = (x³ - 7x² + 12x - 10)

We need to follow the above similar steps to find for which value of x, g(x) = 0.

Applying hit and trail method. Checking for x = ±1, ±2, ±3, ±4, ±5... etc.

We find that for x = 5, g(x) = 0.

so x = 5 is a root for the polynomial f(x) = x³ - 7x² + 12x - 10.

=> (x - 5 = 0) is a root for the polynomial f(x) = x³ - 7x² + 12x - 10.

Now factorizing g(x) = x³ - 7x² + 12x - 10, taking (x - 5) as common factor , we get:

g(x) = x³ - 5x² - 2x² + 10x + 2x - 10

g(x) = x2(x - 5) - 2x(x - 5) + 2(x - 5)

g(x) = (x - 5)(x² - 2x + 2)

Now the final step we need to factorize this part x² - 2x + 2.

We find it's difficult to find the root using normal vanishing factor procedure.

Applying Sridharacharya's theorem we know if a quadratic polynomial is in the form ax² + bx + c = 0, then the roots are given as:

x = (-b ± √(b² - 4ac)) / 2a

So here b = -2, a = 1, c = 2

Therefore, x = (-(-2) ± √((-2)² - 4×1×2)) / (2 × 1)

x = (2 ± √(-4)) / 2

x = (2 ± 2i) / 2 [√(-1) = i]

x = 1 ± i

x = 1 + i or x = 1 - i

x - (1 + i) = 0 or x - (1 - i) = 0

(x - (1 + i))(x - (1 - i)) = 0

Finally f(x) = x⁴ - 4x³ - 9x² + 26x - 30 can be written as:

f(x) = (x + 3)(x - 5)(x - (1 + i))(x - (1 - i))

Hence the roots of f(x) or values of x are -3, 5, (1 + i), (1 - i)

Know more about "roots of a polynomial equation" here: https://brainly.com/question/13006165

#SPJ9