Question

Question for maths lovers.
Factorise the given Quintic Polynomial.
$\sf 2 {x}^{5} - {x}^{4} + 10 {x}^{3} - 5 {x}^{2} + 8x - 4$
Solve it if you can.
Moderators can answer in my question.​

Answer 1

Hope, this would be the answer.

Answer 2

Factoring: 2x5+x4-10x3-5x2+8x+4

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

Group 1: -10x3-5x2

Group 2: 2x5+x4

Group 3: 8x+4

Pull out from each group separately :

Group 1: (2x+1) • (-5x2)

Group 2: (2x+1) • (x4)

Group 3: (2x+1) • (4)

Add 1+2+3 :Factoring x4-5x2+4

The first term is, x4 its coefficient is 1 .

The middle term is, -5x2 its coefficient is -5 .

The last term, "the constant", is +4

Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is -5 .

-4 + -1 = -5 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and -1

x4 - 4x2 - 1x2 - 4

Step-4 : Add up the first 2 terms, pulling out like factors :

x2 • (x2-4)

Add up the last 2 terms, pulling out common factors :

1 • (x2-4)

Step-5 : Add up the four terms of step 4 :

(x2-1) • (x2-4)

Which is the desired factorization

Factoring: x2-1

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 : x2 is the square of x1

Factorization is : (x + 1) • (x - 1)

(x+1)•(x-1)•(x+2)•(x-2)•(2x+1)