Question

factorize this expression k3 -4k-12​

Answer 1

Step by Step Solution

More Icon

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "k2" was replaced by "k^2". 1 more similar replacement(s).

STEP

1

:

Checking for a perfect cube

1.1 k3-k2-k+1 is not a perfect cube

Trying to factor by pulling out :

1.2 Factoring: k3-k2-k+1

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

Group 1: -k+1

Group 2: k3-k2

Pull out from each group separately :

Group 1: (-k+1) • (1) = (k-1) • (-1)

Group 2: (k-1) • (k2)

-------------------

Add up the two groups :

(k-1) • (k2-1)

Which is the desired factorization

Trying to factor as a Difference of Squares:

1.3 Factoring: k2-1

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 : 1 is the square of 1

Check : k2 is the square of k1

Factorization is : (k + 1) • (k - 1)

Multiplying Exponential Expressions:

1.4 Multiply (k - 1) by (k - 1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (k-1) and the exponents are :

1 , as (k-1) is the same number as (k-1)1

and 1 , as (k-1) is the same number as (k-1)1

The product is therefore, (k-1)(1+1) = (k-1)2

Final result :

(k + 1) • (k - 1)2