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One root of f(x)=x^3-4x^2-20x+48 is x = 6. What are all the factors of the function? Use the Remainder Theorem.


a.(x + 6)(x + 8)

b.(x – 6)(x – 8)

c.(x – 2)(x + 4)(x – 6)

d.(x + 2)(x – 4)(x + 6)

Verified Answer :

HomertheGenius

If one root is x = 6, then this polynomial is divisible by: x - 6.
( x³ - 4 x² - 20 x + 48 ) : ( x - 6 ) = x² + 2 x - 8
- x³ + 6x²
---------------
         2 x² - 20 x
        -2x² + 12 x
       -----------------
                 - 8 x + 48
                   8 x - 48
               --------------------
After that : x²+ 2 x - 8 = x² + 4 x - 2 x - 8 = x ( x + 4 ) - 2 ( x + 4 ) =
= ( x + 4 ) ( x - 2 )
Answer: C )  ( x - 2 ) ( x + 4 ) ( x - 6 )

abidemiokin

The factored form of the polynomial is (x-6)(x-2)(x+4)

Polynomial factorization

Given the polynomial function expreessed as g(x) = x^3-4x^2-20x+48

Group the functions

g(x) =  x^3-4x^2-20x+48/x-6 = x² + 2 x - 8

Factorize the quotient

x² + 2 x - 8

x^2 + 4x - 2x - 8  =0

x(x+4)-2(x+4) = 0

(x-2)(x+4)

Hence the factored form of the polynomial is (x-6)(x-2)(x+4)

Learn more on polynomial  heree: https://brainly.com/question/2833285

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