( 7 x 10 ^-5) x (5 x 10 ^-8)

Step 1 : Equation at the end of step 1

(((7•(x^10 ))-5)•x)•(5x^10-8)

Step 2 : Equation at the end of step 2 :

((7x^10 - 5) • x) • (5x^10 - 8)

Step 3 :

Trying to factor as a Difference of Squares:

3.1 Factoring: 7x^10-5

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 =

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

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

Check : 7 is not a square !!

Ruling : Binomial can not be factored as the

difference of two perfect squares.

Equation at the end of step 3 :

x • (7x10 - 5) • (5x10 - 8)

Step 4 :

4.1 Factoring: 5x^10-8

Check : 5 is not a square !!

Ruling : Binomial can not be factored as the

difference of two perfect squares.

Final result : ☞ x • (7x10 - 5) • (5x10 - 8)

→ 3.5 × 10^-12

____________________________

Hope It Helps You ✌️

Step 1 : Equation at the end of step 1

(((7•(x^10 ))-5)•x)•(5x^10-8)

Step 2 : Equation at the end of step 2 :

((7x^10 - 5) • x) • (5x^10 - 8)

Step 3 :

Trying to factor as a Difference of Squares:

3.1 Factoring: 7x^10-5

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 =

\: \: \: \: \: \blue {A² - 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 : 7 is not a square !!

Ruling : Binomial can not be factored as the

difference of two perfect squares.

Equation at the end of step 3 :

x • (7x10 - 5) • (5x10 - 8)

Step 4 :

4.1 Factoring: 5x^10-8

Check : 5 is not a square !!

Ruling : Binomial can not be factored as the

difference of two perfect squares.

Final result : ☞ x • (7x10 - 5) • (5x10 - 8)

→ 3.5 × 10^-12

____________________________

Hope It Helps You ✌️