How many roots of the equation `3x^4+6x^3+x^2+6x+3=0` are real ?

Answer

Two

Explanation

Let `f(x) = 3x^(4) + 6x ^(3) + x^(2) + 6x + 3`
` therefore f' (x) = 12 x^(3) + 18x^(2) + 2x + 6`
and `f''(x) = 36x^(2) +36x + 2x`
`= 2 (18x^(2) + 18x + 2) ne`o for any real x.
So, f(x) = 0 has maximum two real roots.
Now ,f(0) = 3 `gt 0 and f( - 1) = 3 - 6 + 1 - 6 + 3 lt 0`
S0, graphe cuts x-axis between -1 and 1 .
Hence, f(x) = 0 has two real roots.