# Factorization of Quadratic Equations

*Every* quadratic equation has two roots. This is a fact which will be justified later.

Consider a quadratic equation \(Q\left( x \right) = 0\). Suppose that \(x = \alpha \) is one root of this equation. This means that \(x = \alpha \) is a zero of the quadratic expression \(Q\left( x \right)\). Thus, \(\left( {x - \alpha } \right)\) should be a factor of \(Q\left( x \right)\). This is an extremely important point, so make sure you understand it (recall the *factor theorem* from an earlier class).

Similarly, if \(x = \beta \) is the second root of \(Q\left( x \right) = 0\), then \(x = \beta \) is a zero of \(Q\left( x \right)\), which means that \(\left( {x - \beta } \right)\) is a factor of \(Q\left( x \right)\).

We thus see that \(Q\left( x \right)\) has two factors: \(\left( {x - \alpha } \right)\) and \(\left( {x - \beta } \right)\), and hence it can be expressed as their product:

\[Q\left( x \right) = k\left( {x - \alpha } \right)\left( {x - \beta } \right)\]

Can you understand why there is a constant factor of \(k\)?

Let us see an example of factorization in terms of the roots. Consider the quadratic expression \(Q\left( x \right)\,\,:\,\,2{x^2} + 5x + 2\), and the corresponding quadratic equation:

\[2{x^2} + 5x + 2 = 0\]

The roots of this equation are \(x = - \frac{1}{2}\) and \(x = 2\). Thus, the two (linear) factors of \(Q\left( x \right)\) will be:

**Factor-1:** \(\left( {x - \left( { - \frac{1}{2}} \right)} \right)\) or \(\left( {x + \frac{1}{2}} \right)\)

**Factor-2:** \(\left( {x - \left( { - 2} \right)} \right)\) or \(\left( {x + 2} \right)\)

We can now express \(Q\left( x \right)\) in terms of the product of these two factors:

\[Q\left( x \right) = 2\left( {x + \frac{1}{2}} \right)\left( {x + 2} \right)\]

Verify that this is indeed correct. Also, note that a constant factor of 2 is required to equalize the coefficients on both sides.

This discussions indicates that to find the roots of a quadratic equation \(Q\left( x \right) = 0\), we can try to factorize \(Q\left( x \right)\) as a product of two linear factors. For example, consider the equation \({x^2} - 3x + 2 = 0\). The expression \({x^2} - 3x + 2\) can be factorized as \(\left( {x - 1} \right)\left( {x - 2} \right)\), so the equation can be written as

\[\left( {x - 1} \right)\left( {x - 2} \right) = 0\]

Thus, the two roots of the equation are \(x = 1\) and \(x = 2\).

Factorization of quadratic expressions can be done by *splitting the middle term*, a technique which you have learnt earlier. Here are some examples:

**Solved Example 1:** Find the values of \(x\) which satisfy

\[6{x^2} - x - 2 = 0\]

**Solution:** We have:

\[\begin{align}&6{x^2} - 4x + 3x - 2 = 0\\&\Rightarrow \,\,\,2x\left( {3x - 2} \right) + 1\left( {3x - 2} \right) = 0\\&\Rightarrow \,\,\,\left( {2x + 1} \right)\left( {3x - 2} \right) = 0\\&\Rightarrow \,\,\,x = - \frac{1}{2},\,\,\,x = \frac{2}{3} \end{align}\]

Note how we split the middle term \( - x\) into \( - 4x + 3x\), which enabled us to carry out the factorization.

**Solved Example 2:**** **Find the values of \(x\) which satisfy

\[4\sqrt 3 {x^2} + 5x - 2\sqrt 3 = 0\]

**Solution:** Factorizing might seem difficult in this case but we note that \(4\sqrt 3 \times \left( { - 2\sqrt 3 } \right) = - 24\). Also, \(8 \times \left( { - 3} \right)\) equals \( - 24\)and \(8 + \left( { - 3} \right) = 5\). Thus, we can split the middle term as follows:

\[\begin{align}&4\sqrt 3 {x^2} + 5x - 2\sqrt 3 = 0\\&\Rightarrow \,\,\,4\sqrt 3 {x^2} + 8x - 3x - 2\sqrt 3 = 0\\&\Rightarrow \,\,\,4x\left( {\sqrt 3 x + 2} \right) - \sqrt 3 \left( {\sqrt 3 x + 2} \right) = 0\\&\Rightarrow \,\,\,\left( {4x - \sqrt 3 } \right)\left( {\sqrt 3 x + 2} \right) = 0\\&\Rightarrow \,\,\,x = \frac{{\sqrt 3 }}{4},\,\,\,x = - \frac{2}{{\sqrt 3 }}\end{align}\]

**Solved Example 3:** Write a quadratic equation whose roots are \(x = \sqrt 2 \) and \(x = \sqrt 3 \).

**Solution:** The following equation will work:

\[\begin{align}&\left( {x - \sqrt 2 } \right)\left( {x - \sqrt 3 } \right) = 0\\&\Rightarrow \,\,\,{x^2} - \left( {\sqrt 2 + \sqrt 3 } \right)x + \sqrt 6 = 0\end{align}\]

**Solved Example 4:** Write a quadratic equation whose roots are \(x = \pi \) and \(x = - 2\) and in which the coefficient of \({x^2}\) is \(\sqrt 2 \).

**Solution:** An equation whose roots are \(x = \pi \) and \(x = - 2\) can be written as follows:

\[\begin{align}&\left( {x - \pi } \right)\left( {x - \left( { - 2} \right)} \right) = 0\\&\Rightarrow \,\,\,\left( {x - \pi } \right)\left( {x + 2} \right) = 0\\&\Rightarrow \,\,\,{x^2} + \left( {2 - \pi } \right)x - 2\pi = 0\end{align}\]

However, we want the coefficient of \({x^2}\) to be \(\sqrt 2 \). We can now multiply the above equation on both sides by \(\sqrt 2 \) (note that this does not change the roots). Thus, the required equation is:

\[\begin{align}&\sqrt 2 \left( {\,{x^2} + \left( {2 - \pi } \right)x - 2\pi } \right) = 0\\& \Rightarrow \,\,\,\sqrt 2 {x^2} + \sqrt 2 \left( {2 - \pi } \right)x - 2\sqrt 2 \pi = 0\end{align}\]

**Solved Example 5:** Solve the following equation:

\[\frac{1}{a} + \frac{1}{b} + \frac{1}{x} = \frac{1}{{a + b + x}}, \,\,\,a + b \ne 0\]

**Solution:** Before solving this equation, can you understand why \(a + b \ne 0\) is an important piece of information?

Combining the terms on the left side, we have:

\[\begin{align}&\frac{{bx + ax + ab}}{{abx}} = \frac{1}{{a + b + x}}\\&\Rightarrow \,\,\,\left\{ {\left( {a + b} \right)x + ab} \right\}\left\{ {\left( {a + b} \right) + x} \right\} = abx\\&\Rightarrow \,\,\,{\left( {a + b} \right)^2}x + \left( {a + b} \right){x^2} + ab\left( {a + b} \right) = 0\\&\Rightarrow \,\,\,{x^2} + \left( {a + b} \right)x + ab = 0\\&\Rightarrow \,\,\,\left( {x + a} \right)\left( {x + b} \right) = 0\end{align}\]

Clearly, the two roots are \( - a,\,\, - b\). Verify by substitution!

**Solved Example 6:** Solve the following equation:

\[\frac{2}{x} + \frac{5}{{x + 2}} = \frac{9}{{x + 4}}\]

**Solution:** First, combine the two terms on the left side:

\[\begin{align}&\frac{{2\left( {x + 2} \right) + 5x}}{{x\left( {x + 2} \right)}} = \frac{9}{{x + 4}}\\&\Rightarrow \,\,\,\frac{{7x + 4}}{{{x^2} + 2x}} = \frac{9}{{x + 4}}\end{align}\]

Now, cross-multiply and rearrange the terms to get a quadratic equation:

\[\begin{align}&\left( {7x + 4} \right)\left( {x + 4} \right) = 9\left( {{x^2} + 2x} \right)\\&\Rightarrow \,\,\,7{x^2} + 32x + 16 = 9{x^2} + 18x\\&\Rightarrow \,\,\,2{x^2} - 14x - 16 = 0\\&\Rightarrow \,\,\,{x^2} - 7x - 8 = 0\\&\Rightarrow \,\,\,\left( {x - 8} \right)\left( {x + 1} \right) = 0\end{align}\]

The two roots are \(8,\,\, - 1\). Verify by substitution that these values of *x* indeed satisfy the original equation.

**Solved Example 7:** If the roots of the equation \(\left( {x - a} \right)\left( {x - b} \right) - p = 0\) are \(c,\,\,d\), find the roots of \(\left( {x - c} \right)\left( {x - d} \right) + p = 0\).

**Solution:** Since the roots of \(\left( {x - a} \right)\left( {x - b} \right) - p = 0\) are \(c,\,\,d\), we can write:

\[\begin{align}&\left( {x - a} \right)\left( {x - b} \right) - p = \left( {x - c} \right)\left( {x - d} \right)\\&\Rightarrow \,\,\,\left( {x - c} \right)\left( {x - d} \right) + p = \left( {x - a} \right)\left( {x - b} \right)\end{align}\]

Clearly, the roots of \(\left( {x - c} \right)\left( {x - d} \right) + p = 0\) are \(a,\,\,b\). This is a very elegant solution! Make sure you pause for a moment and appreciate its significance.

**Solved Example 8:** Solve the following equation:

\[2\left( {\frac{{2x - 1}}{{x + 3}}} \right) - 3\left( {\frac{{x + 3}}{{2x - 1}}} \right) = 5,\,\,\,\,\,x \ne \left( { - 3,\,\,\frac{1}{2}} \right)\]

**Solution:** Note that on the left side above, the expressions in the two brackets are reciprocals of each other. We make the following substitution:

\[\frac{{2x - 1}}{{x + 3}} = y\]

This reduces the given equation to:

\[\begin{align}&2y - \frac{3}{y} = 5\\&\Rightarrow \,\,\,2{y^2} - 5y - 3 = 0\end{align}\]

We can easily solve this by factorization:

\[\begin{align}&2{y^2} - 6y + y - 3 = 0\\ &\Rightarrow \,\,\,2y\left( {y - 3} \right) + 1\left( {y - 3} \right) = 0\\& \Rightarrow \,\,\,\left( {2y + 1} \right)\left( {y - 3} \right) = 0\end{align}\]

We now have to take two cases, corresponding to the two roots:

### Case-1

\[\begin{align}&y = - \frac{1}{2}\,\,\, \Rightarrow \,\,\,\frac{{2x - 1}}{{x + 3}} = - \frac{1}{2}\\&4x - 2 = - x - 3\\&\Rightarrow \,\,\,5x = - 1\\&\Rightarrow \,\,\,x = - \frac{1}{5}\end{align}\]

### Case-2

\[\begin{align}&y = 3\,\,\,\, \Rightarrow \,\,\,\frac{{2x - 1}}{{x + 3}} = 3\\&\Rightarrow \,\,\,2x - 1 = 3x + 9\\&\Rightarrow \,\,\,x = - 10\end{align}\]

Thus, the two roots of the original equation are \( - \frac{1}{5},\,\, - 10\).