Question

Factorize completely a2- 16 b2 - 2a -8b​

Answer 1

Answer:

−2⋅(a 3 −16ab 2−a+5b)

Step-by-step explanation:

See steps

Step by Step Solution:

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Reformatting the input :

Changes made to your input should not affect the solution:

(1): "b2" was replaced by "b^2". 1 more similar replacement(s).

STEP

1

:

Equation at the end of step 1

((a2)-24b2)

((2a-7b)-((4•———————————)•a))-3b

2

STEP

2

:

a2 - 16b2

Simplify —————————

2

Trying to factor as a Difference of Squares:

2.1 Factoring: a2 - 16b2

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 : 16 is the square of 4

Check : a2 is the square of a1

Check : b2 is the square of b1

Factorization is : (a + 4b) • (a - 4b)

Equation at the end of step

2

:

(a+4b)•(a-4b)

((2a-7b)-((4•—————————————)•a))-3b

2

STEP

3

:

Equation at the end of step 3

((2a-7b)-(2•(a+4b)•(a-4b)•a))-3b

STEP

4

:

Equation at the end of step 4

((2a - 7b) - 2a • (a + 4b) • (a - 4b)) - 3b

STEP

5

:

STEP

6

:

Pulling out like terms

6.1 Pull out like factors :

-2a3 + 32ab2 + 2a - 10b =

-2 • (a3 - 16ab2 - a + 5b)

Checking for a perfect cube :

6.2 a3 - 16ab2 - a + 5b is not a perfect cube

Final result :

-2 • (a3 - 16ab2 - a + 5b)