1. Using Basic Algebra

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


  1. By Guessing

Example: x3+2x2−x

x3+2x2−x = x(x2+2x−1)

Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.


We graph the polynomial and check where it crosses the x-axis.

Graphing is a good way to find approximate answers, or sometimes may be an exact answer.



When a polynomial is factored in the form of:


Then, a, b, c etc. are the roots.



we see (x-2) 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) = (23+3.22)(2-2)

f(2) = (23+3.22)(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) = ax4 + bx3 + …

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)




ax2 + bx + c

When the roots are p and q, the same quadratic becomes:



Let’s expand


= a(x2 − px − qx + pq)

= ax2 − a(p+q)x + apq


Now let us compare:

Quadratic: ax2 +bx +c
Expanded Factors: ax2 −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


Sum of the roots = −b/a

Product of the roots = c/a


Finding the quadratic equation:

x2 − (sum of the roots)x + (product of the roots) = 0




Now let us look at the following:

ax3 + bx2 + cx + d

Let us expand the factors:


= ax3 − a(p+q+r)x2 + a(pq+pr+qr)x − a(pqr)


Again comapring:

Cubic: ax3 +bx2 +cx +d
Expanded Factors: ax3 −a(p+q+r)x2 +a(pq+pr+qr)x −apqr


We can now see that −a(p+q+r)x2 = bx2, 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


Please enter your comment!
Please enter your name here