Question

WILL MARK BRAINLIEST consider the polynomial expression below x^3-3x^2+81x-243 rewrite the polynomial in the form (x-d)(x-e)(x+f) where d is a real number, and e and f are complex numbers in the form bi

Answer 1

Answer:

x³ - 3x² + 81x - 243  = ( x - 3 )( x - 9i )( x + 9i )

Step-by-step explanation:

Given Expression is x³ - 3x² + 81x - 243

We need to factorize the given expression in form of ( x - d )( x - e )( x + f ) where d is a real number, and e and f are complex numbers in the form bi.

From Rational factor theorem,

Factors of 243 is ±1 , ±3 , ±9 , ±27 , ±81 , ±243.

Now by hit and trial we find first factor of real number.

let, p(x) = x³ - 3x² + 81x - 243

put x = 3

p(3) = 3³ - 3 × 3² + 81 × 3 - 243  = 27 - 27 + 243 - 243 = 0

So, The first factor is ( x - 3 ).

Now to find other factor we divide given polynomial with x - 3.

Division is attached.

We get Quotient = x² + 81

Remainder = 0

Now, Factorize x² + 81 = ( x )² - ( 9i )² = ( x - 9i )( x + 9i )

Therefore, x³ - 3x² + 81x - 243  = ( x - 3 )( x - 9i )( x + 9i )

Answer 2

Answer:

(x - 3)(x - 9i)(x + 9i)

Step-by-step explanation:

note that x = 3 gives

3³ - 3(3)² + 81(3) - 243 = 27 - 27 + 243 - 243 = 0, hence

x = 3 is a root and (x - 3) is a factor

x³ - 3x² + 81x - 243 ÷ (x - 3)

= (x - 3)(x² + 81)

solve x² + 81 = 0 ⇒ x² = - 81 ⇒ x = ± 9i, thus

x³ - 3x² + 81x - 243 = (x - 3)(x - 9i)(x + 9i) ← in factored form