3x+1 is a linear polynomial. The graph of y = 2x+1 is a straight line.
A "root" is when y is zero: 3x+1 = 0
Subtract 1 from both sides: 3x = −1
Divide both sides by 3: x = −1/3
And that is the solution:
x = −1/3
Example: x^{3}+2x^{2}−x
x^{3}+2x^{2}−x = x(x^{2}+2x−1)
Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.
2. Graphically.
We graph the polynomial and check where it crosses the xaxis.
Graphing is a good way to find approximate answers, or sometimes maybe an exact answer.
When a polynomial is factored in the form of:
f(x)=(xa)(xb)(xc)….
Then, a, b, c etc. are the roots.
Example,
f(x)=(x^{3}+3x^{2})(x2)
we see (x2) and that means 2 is the root (or zero) of the function.
We can check this if we put x=2 in the function.
f(2) = (2^{3}+3.2^{2})(22)
f(2) = (2^{3}+3.2^{2})(0)
f(3) = 0, so 3 is a rot of the function.
How to check if we have found the root (or zero) of the polynomial?
Simply put the root in place of "x": the polynomial should be equal to zero.
Number of roots = the degree of polynomial
Degree 
Roots 
Possible Combinations 
1 
1 
1 Real Root 
2 
2 
2 Real Roots, or 2 Imaginary Roots 
3 
3 
3 Real Roots, or 1 Real and 2 Imaginary Roots 
4 
4 
4 Real Roots, or 2 Real and 2 Imaginary Roots, or 4 Imaginary Roots 
etc 

etc! 
We can take a polynomial, such as:
f(x) = ax^{4} + bx^{3} + ...
And then factor it like this:
f(x) = a(x−p)(x−q)(x−r)...
Then p, q, r, etc are the roots (where the polynomial equals zero)
ax^{2} + bx + c
When the roots are p and q, the same quadratic becomes:
a(x−p)(x−q)
Let's expand
a(x−p)(x−q)
= a(x^{2} − px − qx + pq)
= ax^{2} − a(p+q)x + apq
Now let us compare:
Quadratic: 
ax^{2} 
+bx 
+c 
Expanded Factors: 
ax^{2} 
−a(p+q)x 
+apq 
We can now see that −a(p+q)x = bx, so:
−a(p+q) = b
p+q = −b/a
And apq = c, so:
pq = c/a
So,
Sum of the roots = −b/a
Product of the roots = c/a
Finding the quadratic equation:
x^{2} − (sum of the roots)x + (product of the roots) = 0
Now let us look at the following:
ax^{3} + bx^{2} + cx + d
Let us expand the factors:
a(x−p)(x−q)(x−r)
= ax^{3} − a(p+q+r)x^{2} + a(pq+pr+qr)x − a(pqr)
Again comapring:
Cubic: 
ax^{3} 
+bx^{2} 
+cx 
+d 
Expanded Factors: 
ax^{3} 
−a(p+q+r)x^{2} 
+a(pq+pr+qr)x 
−apqr 
We can now see that −a(p+q+r)x^{2} = bx^{2}, so:
−a(p+q+r) = b
p+q+r = −b/a
And −apqr = d, so:
pqr = −d/a
Adding the roots gives −b/a
Multiplying the roots gives −d/a
Also, pq+pr+qr = c/a
Hope now you have understood how to solve polynomials. Share your experience with us in the comment section below.