Question

Factorise ${{x}^{2}}+5x+6$ into linear factors.

Answer 1

Hint: Use $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$ or factorise using the method of completing the square or use quadratic formula [$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$] and factor theorem [if $f(a)=0$ then $x-a$ is the factor of f(x)].

Complete step-by-step solution -
Let ${{x}^{2}}+5x+6=\left( x+a \right)\left( x+b \right)$
Using $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$ we get
${{x}^{2}}+5x+6={{x}^{2}}+\left( a+b \right)x+ab$
Comparing coefficients of x, we get
$a+b=5\text{ (i)}$
Comparing constant terms, we get
$ab=6\text{ (ii)}$
We can factorise 6 in following ways
$6=6\times 1$ or $6=3\times 2$
If a = 6 and b = 1 then we have $a+b=7\ne 5$
If a = 3 and b = 2 then we have $a+b=3+2=5$
Hence, we get a = 3 and b = 2
$\Rightarrow {{x}^{2}}+5x+6=\left( x+3 \right)\left( x+2 \right)$
Sometimes it is not possible to factorise by using hit and trial method as above. In those cases, we factorise using the method of completing the square.
In this method we use the following identities to factorise
[i] ${{x}^{2}}+2ax+{{a}^{2}}={{\left( x+a \right)}^{2}}$
[ii] ${{x}^{2}}-2ax+{{a}^{2}}={{\left( x-a \right)}^{2}}$
[iii] ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Step 1: Make the coefficient of leading term as 1
Here the leading term has a coefficient of the leading term is already one. So, we proceed to step 2.
Step 2: Write the middle term in the form of 2ax.
${{x}^{2}}+5x+6={{x}^{2}}+2\times \dfrac{5}{2}\times x+6$
Here the middle term 5x has been written in form of 2ax with a = $\dfrac{5}{2}$
Step 3: Add and Subtract ${{a}^{2}}$
\[\Rightarrow {{x}^{2}}+5x+6={{x}^{2}}+2\times \dfrac{5}{2}x+6+{{\left( \dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{5}{2} \right)}^{2}}\]
Step 4: Use ${{x}^{2}}+2ax+{{a}^{2}}={{\left( x+a \right)}^{2}}$ or ${{x}^{2}}-2ax+{{a}^{2}}={{\left( x-a \right)}^{2}}$ whichever is applicable
${{x}^{2}}+2\times \dfrac{5}{2}x+{{\left( \dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{5}{2} \right)}^{2}}+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{5}{2} \right)}^{2}}+6$ using ${{x}^{2}}+2ax+{{a}^{2}}={{\left( x+a \right)}^{2}}$
$\Rightarrow {{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{5}{2} \right)}^{2}}+6$
Simplifying we get
$\begin{align}
  & \Rightarrow {{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-\dfrac{25}{4}+6 \\
 & \Rightarrow {{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-\dfrac{25-6\times 4}{4} \\
 & \Rightarrow {{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-\dfrac{25-24}{4} \\
 & \Rightarrow {{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-\dfrac{1}{4} \\
\end{align}$
Step 5: Write the constant term in the form of ${{b}^{2}}$
$\Rightarrow {{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-\dfrac{1}{4}={{\left( x+\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}$
Step 6: If the resultant expression is of form ${{a}^{2}}+{{b}^{2}}$then the given expression cannot be factorised.
If it is of the form ${{a}^{2}}-{{b}^{2}}$ then use ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Here the resultant expression is of the form of ${{a}^{2}}-{{b}^{2}}$ where $a=\left( x+\dfrac{5}{2} \right)$ and \[b\text{ }=\dfrac{1}{2}\]
Using ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we get
${{x}^{2}}+5x+6={{\left( x+\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}=\left( x+\dfrac{5}{2}+\dfrac{1}{2} \right)\left( x+\dfrac{5}{2}-\dfrac{1}{2} \right)$
Simplifying we get

$\begin{align}
  & {{x}^{2}}+5x+6=\left( x+\dfrac{5}{2}+\dfrac{1}{2} \right)\left( x+\dfrac{5}{2}-\dfrac{1}{2} \right)=\left( x+\dfrac{5+1}{2} \right)\left( x+\dfrac{5-1}{2} \right) \\
 & \Rightarrow {{x}^{2}}+5x+6=\left( x+\dfrac{6}{2} \right)\left( x+\dfrac{4}{2} \right)=\left( x+3 \right)\left( x+2 \right) \\
\end{align}$
Hence ${{x}^{2}}+5x+6=\left( x+3 \right)\left( x+2 \right)$

Note: Quadratic formula: The roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are $\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ and }\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$.
We can factorise quadratic expressions using quadratic formulas also.
If ${{r}_{1}}$ and ${{r}_{2}}$ are the roots of quadratic expression(found using quadratic formula) then the expression = $a\left( x-{{r}_{1}} \right)\left( x-{{r}_{2}} \right)$ where a is the leading coefficient of the quadratic expression.
Here a = 1, b = 5 and c = 6
We know from quadratic formula ${{r}_{1}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and ${{r}_{2}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of a, b and c in the expression for ${{r}_{1}}$ and ${{r}_{2}}$ we get
${{r}_{1}}=\dfrac{-5+\sqrt{{{5}^{2}}-4\times 1\times 6}}{2\times 1}$ and ${{r}_{2}}=\dfrac{-5-\sqrt{{{5}^{2}}-4\times 1\times 6}}{2\times 1}$
$\Rightarrow {{r}_{1}}=-2\text{ and }{{r}_{2}}=-3$
Hence ${{x}^{2}}+5x+6=\left( x-\left( -2 \right) \right)\left( x-\left( -3 \right) \right)$
$\Rightarrow {{x}^{2}}+5x+6=\left( x+2 \right)\left( x+3 \right)$