Question

What are the real zeroes of x^3+ 6x^2- 9x - 54?

Answer 1

Answer:

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Answer 2

Answer:

(x + 3) • (x - 3) • (x + 6)

Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

 (((x3) +  (2•3x2)) -  9x) -  54

Step  2  :

Checking for a perfect cube :

2.1    x3+6x2-9x-54  is not a perfect cube

Trying to factor by pulling out :

2.2      Factoring:  x3+6x2-9x-54

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -9x-54

Group 2:  6x2+x3

Pull out from each group separately :

Group 1:   (x+6) • (-9)

Group 2:   (x+6) • (x2)

              -------------------

Add up the two groups :

              (x+6)  •  (x2-9)

Which is the desired factorization

Trying to factor as a Difference of Squares :

2.3      Factoring:  x2-9

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 9 is the square of 3

Check :  x2  is the square of  x1

Factorization is :       (x + 3)  •  (x - 3)

Final result :

 (x + 3) • (x - 3) • (x + 6)

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