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: 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.
2. Graphically.
We graph the polynomial and check where it crosses the x-axis.
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)=(x-a)(x-b)(x-c)….
Then, a, b, c etc. are the roots.
Example,
f(x)=(x3+3x2)(x-2)
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:
a(x−p)(x−q)
Let's expand
a(x−p)(x−q)
= 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
So,
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:
a(x−p)(x−q)(x−r)
= 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
Hope now you have understood how to solve polynomials. Share your experience with us in the comment section below.