Finding Roots

Video lesson · from 32:001 worked example

One lesson video covers all of Polynomials & Quadratics, so it opens at 32:00 for this topic — not from the beginning.

Theory

Finding the roots of a polynomial equation $f(x)=0$ follows the same principles as finding roots of a quadratic. We must first fully factorise the polynomial and then set each factor equal to zero.

⚠️ Common Examiner Traps

Factorise fully before solving: a partly factorised polynomial hides roots. Keep going until every factor is linear or an irreducible quadratic.
Reverse the sign for the root: the factor $(x+4)$ gives the root $x = -4$ .
Check the quadratic factor as well: it may factorise further, need the quadratic formula, or have no real roots — use the discriminant to decide.
Say what you have found: state the roots clearly at the end rather than leaving them inside brackets.

Worked examples

Example 1

Given that $f(x)=x^3-37x+84$ ,

a) Show that $x=-7$ is a root of $f(x)=0$

b) Solve $f(x)=0$ .

Substitute $x = -7$ into the polynomial (note there is no $x^2$ term, so its coefficient is 0):

f(-7) = (-7)^3 - 37(-7) + 84

f(-7) = -343 + 259 + 84 = 0

Since $f(-7) = 0$ , $x = -7$ is a root.

Since $x=-7$ is a root, $(x+7)$ is a factor.

We use synthetic division to divide $x^3-37x+84$ by $(x+7)$ to find the quotient.

-7	1	0	-37	84
	-7	49	-84
	1	-7	12	0

The quotient is $x^2-7x+12$ .

x^3-37x+84 = (x+7)(x^2-7x+12)

Factorise the quadratic $x^2-7x+12$ :

x^2-7x+12 = (x-3)(x-4)

Set the fully factorised polynomial to 0:

(x+7)(x-3)(x-4) = 0

Therefore, the roots are $x = -7, x = 3, x = 4$ .