**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*

**By Guessing**

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.

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.

# Factors

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)=(x ^{3}+3x^{2})(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) = (2 ^{3}+3.2^{2})(2-2)*

*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)

# Quadratic

*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*

# Cubic

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*