# Factorization

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 $$- \frac{1}{5},\,\, - 10$$.

grade 10 | Questions Set 2
grade 10 | Questions Set 1