Question

solve:

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

Answer 1

$p\left(x\right)=6x^4-25x^3+12x^2+25x+6=0$

Take LHS of above equation & put $x=2$

$6(2{)}^4-25(2{)}^3+12(2{)}^2+25(2)+6$

$6\times 16-25\times 8+12\times 4+75+6$

$96-200+48+56$

$=-200+200=0$

$\Rightarrow (x-2)$ is a factor of $p\left(x\right)$

Now put $x=3$ in $p\left(x\right)$

$6(3{)}^4-25(3{)}^3+12(3{)}^2+25(3)+6$

$6\times 81-25\times 27+12\times 9+25\times 3+6$

$486-675+108+75+6$

$=-675+675=0$

$\Rightarrow (x-3)$ is a factor of $p\left(x\right)$

Now $(x-3)(x-2)=x^2-5x+6$ is a factor of $p\left(x\right)$

$x^2-5x+6)6x^4-25x^3+12x^2+25x+16(6x^2+5x+1$

$6x^4-30x^3+36x^2$

______________________

$5x^3-24x^2+25x+6$

$5x^3-25x^2+30x$

_____________________

$x^2-5x+6$

$x^2-5x+6$

_______________

$0$

_______________

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

$(x-2)(x-3)(6x^2+5x+1)=0$

$(x-2)(x-3)(6x^2+3x+2x+1)=0$

$(x-2)(x-3)\left(3x\right(6x+1)+(2x+1\left)\right)=0$

$(x-2)(x-3)(3x+1)(2x+1)=0$

$\Rightarrow x=2,3,-1/3,-1/2$.