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How to Factorize Quadratic Expressions

You factorize a quadratic expression $f (x)$ by finding its roots. To find those roots, just solve the equation $f (x) = 0$ , which can be done with the help of one of these methods:

1.: By using the quadratic formula;
2.: By completing the square;
3.: Through inspection, which means looking for solutions that fit the expression without doing the entire calculation.

Rule

Factorization of Quadratic Equations

Find the roots of the expression, $x_{1}$ and $x_{2}$ . You can factorize the expression like this:

\begin{matrix} a x^{2} + b x + c \\ = \\ a (x - x_{1}) (x - x_{2}) . \end{matrix}

a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) .

If the equation only has one solution

x_{1}

, that means that

x_{1} = x_{2}

. In that case, the factorized expression will be:

a {(x - x_{1})}^{2} .

If the equation doesn’t have any solutions, the expression can’t be factorized.

Example 1

The Quadratic Formula; Two Real Solutions

Factorize $f (x) = x^{2} + 11 x + 30$ .

Use the quadratic formula. Here, we have $a = 1$ , $b = 11$ and $c = 30$ , which gives us

\begin{array}{l} x & = \frac{- 11 \pm \sqrt{1 1^{2} - 4 \cdot 1 \cdot 30}}{2 \cdot 1} \\ = \frac{- 11 \pm \sqrt{121 - 120}}{2} \\ = \frac{- 11 \pm \sqrt{1}}{2} \\ = \frac{- 11 \pm 1}{2} . \end{array}

This means that

x_{1} = - 5 x_{2} = - 6,

which tells us that the factorized expression is

\begin{matrix} 1 \cdot (x - (- 5)) (x - (- 6)) \\ = \\ (x + 5) (x + 6) . \end{matrix}

1 \cdot (x - (- 5)) (x - (- 6)) = (x + 5) (x + 6) .

Example 2

Complete the Square; Two Real Solutions

Factorize $f (x) = x^{2} - 6 x + 8$ .

\begin{array}{l} x^{2} - 6 x + 8 \\ = x^{2} - 6 x + {(\frac{6}{2})}^{2} - {(\frac{6}{2})}^{2} + 8 \\ = \underset{2nd algebraic identity}{\underset{⏟}{x^{2} - 6 x + 3^{2}}} - 3^{2} + 8 \\ = {(x - 3)}^{2} - 9 + 8 \\ = {(x - 3)}^{2} - 1 \\ = \underset{3rd algebraic identity}{\underset{⏟}{{(x - 3)}^{2} - 1^{2}}} \\ = ((x - 3) - 1) ((x - 3) + 1) \\ = (x - 4) (x - 2) \end{array}

\begin{array}{l} x^{2} - 6 x + 8 & = x^{2} - 6 x + {(\frac{6}{2})}^{2} - {(\frac{6}{2})}^{2} + 8 \\ = \underset{The second algebraic identity}{\underset{⏟}{x^{2} - 6 x + 3^{2}}} - 3^{2} + 8 \\ = {(x - 3)}^{2} - 9 + 8 \\ = {(x - 3)}^{2} - 1 \\ = \underset{The third algebraic identity}{\underset{⏟}{{(x - 3)}^{2} - 1^{2}}} \\ = ((x - 3) - 1) ((x - 3) + 1) \\ = (x - 4) (x - 2) \end{array}

To factorize an expression by completing the square, we have to first use one of the algebraic identities to complete the square, and then the third algebraic identity to finish the factorization.

Example 3

Recognize a Completed Square; One Real Solution

Factorize $f (x) = x^{2} - 2 x + 1$ .

\begin{array}{l} x^{2} - 2 x + 1 & = \underset{2nd algebraic identity}{\underset{⏟}{x^{2} - 2 \cdot 1 \cdot x + 1^{2}}} \\ = {(x - 1)}^{2} \end{array}

x^{2} - 2 x + 1 = \underset{The second algebraic identity}{\underset{⏟}{x^{2} - 2 \cdot 1 \cdot x + 1^{2}}} = {(x - 1)}^{2}

If a quadratic equation only has one real solution, the quadratic expression is a complete square. That’s something we can look for when we attack a new quadratic equation.

Example 4

The Quadratic Equation; One Real Solution

Factorize $f (x) = x^{2} + 8 x + 16$

Use the quadratic formula. Here, we have $(a = 1)$ , $(b = 8)$ and $(c = 16)$ , which gives us

\begin{array}{l} x & = \frac{- 8 \pm \sqrt{8^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \\ = \frac{- 8 \pm \sqrt{64 - 64}}{2} \\ = \frac{- 8 \pm 0}{2} \\ = - 4 . \end{array}

Because we only have one solution, the factorized expression becomes

1 \cdot {(x - (- 4))}^{2} = {(x + 4)}^{2} .

Example 5

The Quadratic Formula; No Real Solutions

Factorize $f (x) = 4 x^{2} + 2 x + 1$

\begin{array}{l} x & = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \\ = \frac{- 2 \pm \sqrt{4 - 16}}{8} \\ = \frac{- 2 \pm \sqrt{- 12}}{8} \end{array}

We have a negative number inside the square root, which means the equation doesn’t have any real solutions. That means you can’t factorize the expression.

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