Question

(a^2+1)(x^2-1)-a^2(a^2+2)x factorise​

Answer 1

hi mate,

solution: (a^2+1)(x^2-1)-a^2(a^2+2)x

Equation at the end of step 1 :

(((a^2)+1)•((x^2)-1))-(a^2•(a^2+2)•x)

Step 2 :

Equation at the end of step 2 :

(((a^2)+1)•((x^2)-1))-a®2x•(a^2+2)

Find roots (zeroes) of : F(a) = a^2+1

Factoring: x^2-1

Theory : A difference of two perfect

squares, A^2 - B^2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A^2 - AB + BA - B^2 =

A^2 - AB + AB - B^2 =

A^2 - B^2

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 : x^2 is the square of x1

Factorization is : (x + 1) • (x - 1)

Equation at the end of step 3 :

(a^2+1)•(x+1)•(x-1)-a^2x•(a^2+2)

Step 4 :

Step 5 :

Pulling out like terms :

5.1 Pull out like factors :

-a^4x + a^2x^2 - 2a^2x - a^2 + x^2 - 1 =

-1 • (a^4x - a^2x^2 + 2a^2x + a^2 - x^2 + 1)

Trying to factor by pulling out :

5.2 Factoring: a^4x - a^2x^2 + 2a^2x + a^2 - x^2 + 1

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

Group 1: 2a^2x + a^2

Group 2: a^4x - a^2x^2

Group 3: -x^2 + 1

Pull out from each group separately :

Group 1: (2x + 1) • (a^2)

Group 2: (a^2 - x) • (a^2x)

Group 3: (-x^2 + 1) • (1) = (x^2 - 1) • (-1)

Looking for common sub-expressions :

Group 1: (2x + 1) • (a^2)

Group 3: (x^2 - 1) • (-1)

Group 2: (a^2 - x) • (a^2x)

Final result :

-a^4x + a^2x^2 - 2a^2x - a^2 + x^2 - 1