Quadratic equations (factorizing)

Summary and examples

A quadratic equation has the following general form:

$ax^2+bx+c=0$

It can be solved by factorization if the following conditions are satisfied:

$a$ , $b$ and $c$ are whole numbers;
$a=1$ ;
if $a\neq1$ and left- and right-hand side are divided by $a$ then the coefficients may not become fractions.

If this method does not work the $abc$ formula has to be applied, see Quadratic equations (abc-formula).

The following examples will clarify this.

Example 1

Is factorizing useful in the following equations, and why (not)?

a. $x^2+5x+6=0$

Yes. $a$ , $b$ and $c$ are whole numbers and $a=1$ .

b. $2x^2+5x+6=0$

No. $a$ , $b$ and $c$ are whole numbers. However, $a\neq1$ and dividing by $2$ gives fractions as coefficients.

c. $2x^2+12x+10=0$

Yes. At first sight you might think 'no', because although $a$ , $b$ and $c$ are whole numbers, $a\neq1$ . However, dividing by $2$ gives the resulting equation:

$x^2+6x+5=0$

which satisfies the conditions. So finally 'yes'.

How do we use factorizing if we want to solve a quadratic equation? The following example will clarify this.

Try to write the equation:

$x^2+5x+6=0$

in the form:

$(x+p)(x+q)=0$

in which $p$ and $q$ are whole numbers.

We have to determine $p$ and $q$ such that the equation:

$(x+p)(x+q)=0$

is equal to the equation:

$x^2+5x+6=0$

This is the case if $p$ and $q$ satisfy the following equations:

$pq=+6$

and:

$p+q=+5$

So find two numbers $p$ and $q$ of which the product equals $6$ and their sum equals $5$ . We find $p=3$ and $q=2$ .

The equation can thus be written as:

$x^2+5x+6=0$

or:

$(x+3)(x+2)=0$

We have to solve the equation:

$(x+3)(x+2)=0$

or:

$x+3=0$ or $x+2=0$

The solution is:

$x=-3$ or $x=-2$

This method is to be preferred when you have enough skills. However, the method is not always applicable and then the $abc$ -formule has to be applied.

Example 2

Solve:

$x^2+6x+5=0$

We have to find $p$ and $q$ such that:

$pq=5$ and $p+q=6$

The following values satisfy:

$p=5$ and $q=1$

and thus the equation can be written as:

$(x+5)(x+1)=0$

The solutions are:

$x=-5$ or $x=-1$

Example 3

Solve:

$x^2+4x-5=0$

We have to find $p$ and $q$ such that:

$pq=-5$ and $p+q=4$

The following values satisfy:

$p=5$ and $q=-1$

Thus the equation can be written as:

$(x+5)(x-1)=0$

with the following solutions:

$x=-5$ or $x=1$

Example 4

Solve:

$x^2-x-12=0$

We have to find $p$ and $q$ such that:

$pq=-12$ and $p+q=-1$

The following values satisfy:

$p=-4$ and $q=3$

Thus the equation can be written as:

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

The solutions are:

$x=4$ or $x=-3$

Example 5

Solve:

$x^2-9x+20=0$

We have to find $p$ and $q$ such that:

$pq=20$ and $p+q=-9$

The following values satisfy:

$p=-5$ and $q=-4$

Thus the equation can be written as:

$(x-5)(x-4)=0$

The solutions are:

$x=5$ or $x=4$

Example 6

Solve:

$x^2-7x-12=0$

We have to find $p$ and $q$ such that:

$pq=-12$ and $p+q=-7$

There are no whole numbers $p$ and $q$ which satisfy these conditions. This does not mean that the quadratic equation has no solutions. In this case it appears that the equation do have solutions and we can find them by applying the $abc$ -formula, see Quadratic equations (abc-formula).

Example 7

Solve:

$x^2-x+20=0$

We have to find $p$ and $q$ such that:

$pq=20$ and $p+q=-1$

There are no whole numbers $p$ and $q$ which satisfy these conditions. This does not mean that the quadratic equation has no solutions. However, in this case it appears that the equation has no solutions, because the discriminant of the equation is negative and then the equation does not have real solutions, see also Quadratic equations (abc-formula).