x^5+2x^4-3x^3-8x^2-4x

Question

Finding zeroes by factorization

Answer 1

Given polynomial x^5+2x^4-3x^3-8x^2-4x

= x(x^4+2x^3-3x^2-8x-4)

= x(x^4-3x^2-4+2x^3-8x) ---> (1)

Now x^4-3x^2-4 = x^4-4x^2+x^2-4

= x^2(x^2-4)+1(x^2-4)

= (x^2+1)(x^2-4)

= (x^2+1)(x+2)(x-2)

Substitute the x^4-3x^2-4 value in (1).

= x[(x+2)(x-2)(x^2+1)+2x^3-8x]

= x[(x+2)(x-2)(x^2+1)+2x(x^2-4)]

= x[(x+2)(x-2)(x^2+1)+2x(x+2)(x-2)]

= x[(x+2)(x-2)((x^2+1)+2x]

= x[(x+2)(x-2)(x^2+2x+1)]

= x[(x+2)(x-2)(x+1)^2]

Zeroes of given polynomial by factoring are x = 0,-1,2,-2.