Question

6x^2+25x^3+12x^2-25x+6

Answer 1

Answer:

Step-by-step explanation:

x= 2, -1/2, 3, -1/3.....!!

6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0

=>6x^4+12x^2+6-25x(x^2-1)=0

=>6(x^2+1)^2 -25x(x^2-1)=0

=>6{(x^2-1)^2 +4x^2} -25x(x^2-1)=0

Let (x^2-1)=a

=>6(a^2+4x^2)-25ax=0

=>6a^2-25ax+24x^2 =0

=>(3a-8x)(2a-3x)=0

=>a=8x/3 or 3x/2!!

Substitute a = x^2-1 and u have 2 quadratics and hence the 4 values which result upon solving the quadratics!!!

Answer 2

Answer:

Step-by-step explanation:

x= 2, -1/2, 3, -1/3.....!!

6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0

=>6x^4+12x^2+6-25x(x^2-1)=0

=>6(x^2+1)^2 -25x(x^2-1)=0

=>6{(x^2-1)^2 +4x^2} -25x(x^2-1)=0

Let (x^2-1)=a

=>6(a^2+4x^2)-25ax=0

=>6a^2-25ax+24x^2 =0

=>(3a-8x)(2a-3x)=0

=>a=8x/3 or 3x/2!!

Substitute a = x^2-1 and u have 2 quadratics and hence the 4 values which result upon solving the quadratics!!!