Magpie Tuition · Higher Maths

Factors and roots

Allow 10–14 minutes. No calculator needed. Before starting: be comfortable with signed terms in brackets (distributing a negative term).

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Overview

Big Picture

A factor is an expression that is multiplied by another expression to make a polynomial. A root is a value of $x$ that makes that polynomial zero.

For a polynomial $f(x)$ :

(x-a)\text{ is a factor of }f(x) \quad\Longleftrightarrow\quad f(a)=0\quad\Longleftrightarrow\quad x=a\text{ is a root of }f(x)=0.

For example, factor $x+3=x-(-3)$ corresponds to root $-3$ . Set the factor equal to zero to find the sign reliably.

Method

Step by Step

For an equation already factorised with zero on one side:

Set each variable factor equal to zero.
Solve each small equation.
List the separate roots using or. Do not add them together.
Check a root by substituting it into the original equation.

Key idea

The Why

The zero-product rule says that if a product equals zero, at least one factor must be zero. Multiplying two non-zero real numbers cannot give zero.

This rule needs a product equal to zero. From $(x-a)(x-b)=12$ , you cannot set either factor equal to zero.

Worked example

Worked examples

1. Equation → factors → roots

Solve $(x-6)(x+3)=0$ .

x-6=0\quad\text{or}\quad x+3=0, \qquad\boxed{x=6\text{ or }x=-3}.

Check both: at $x=6$ , $0\times9=0$ ; at $x=-3$ , $(-9)\times0=0$ . Either value works on its own.

2. Roots → factors → an equation

Write a quadratic equation with roots $-4$ and $7$ .

Root $-4$ gives factor $x-(-4)=x+4$ . Root $7$ gives factor $x-7$ . One suitable equation is

\boxed{(x+4)(x-7)=0}.

In expanded form this is $x^2-7x+4x-28=0$ , or $x^2-3x-28=0$ . You do not need to expand to read its roots. Multiplying the entire equation by a non-zero constant would give the same roots.

Common mistake

The Trap

In example 1, the roots are $6$ and $-3$ , not $-6$ and $3$ . Their sum is $3$ , but that sum is not a replacement for the two roots: $(3-6)(3+3)=-18\ne0$ .

If you found one root, check the other factor. If you found the sum, return to the two separate equations. These checks help locate a possible slip without assuming its cause.

Practice

Your Turn

Practice only — these are not your fresh-check questions. Show the zero-factor equations where relevant.

What root corresponds to each factor: (a) $x-9$ , (b) $x+7$ ?
What linear factor corresponds to each root: (a) $8$ , (b) $-6$ ? Write one quadratic equation having these two roots.
Solve $(x-5)(x+8)=0$ .
Solve $(x+2)(x+9)=0$ .
Given $f(x)=(x-7)(x+4)$ , calculate $f(-4)$ and $f(4)$ . Which value is a root? Name its corresponding linear factor.
A student solves $(x-3)(x+6)=0$ and reports $x=-3$ , because $3+(-6)=-3$ . Explain the error, state the roots and check the reported value.

Answers

Answers with working

Show the answers

(a) $x-9=0\Rightarrow\boxed{x=9}$ . (b) $x+7=0\Rightarrow\boxed{x=-7}$ .
(a) $\boxed{x-8}$ . (b) $x-(-6)=\boxed{x+6}$ . One equation is $\boxed{(x-8)(x+6)=0}$ .
$x-5=0$ or $x+8=0$ , so $\boxed{x=5\text{ or }-8}$ . Checks: $0\times13=0$ and $(-13)\times0=0$ .
$x+2=0$ or $x+9=0$ , so $\boxed{x=-2\text{ or }-9}$ . Checks: $0\times7=0$ and $(-7)\times0=0$ .
$f(-4)=(-4-7)(-4+4)=(-11)\times0=0$ . $f(4)=(4-7)(4+4)=(-3)\times8=-24$ . Thus $\boxed{-4}$ is a root and $\boxed{x+4}$ is its factor; $4$ is not a root.
A root must make the product zero; adding roots is not part of the zero-product rule. $x-3=0$ or $x+6=0$ gives $\boxed{x=3\text{ or }-6}$ . The reported $-3$ gives $(-3-3)(-3+6)=(-6)\times3=-18\ne0$ .

Before moving on: explain the difference between a factor, a root and a sum of roots. Use fresh questions later to check retention; this practice alone does not establish mastery.

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