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Factorise the equation ${x^2} - 9x + 18$.

seo-qna
Last updated date: 24th Jun 2024
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Answer
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Hint: The given expression is ${x^2} - 9x + 18$. We can see that the coefficient of $x$ is $ - 9$ and constant term is $18$. We can express these numbers as sum and product respectively using the numbers six and three. Then we can rearrange the terms to get the factors. Since the given equation is second degree, we will get two linear (first degree) factors.

Complete step-by-step solution:
We are given the equation, ${x^2} - 9x + 18$.
We can split the term $9x$.
Since $9 = 6 + 3$, we have,
${x^2} - 9x + 18 = {x^2} - (6 + 3)x + 18$
Opening brackets we get,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 18$
Now we can see the constant term is $18$.
Also $18 = 6 \times 3$
So we can rewrite the equation as,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 6 \times 3$
Now we can take $x$ common from the first two terms and $ - 3$ common from the last two terms of the right side.
Thus we have,
${x^2} - 9x + 18 = x(x - 6) - 3(x - 6)$
So we get a common factor $x - 6$.
This gives,
${x^2} - 9x + 18 = (x - 6)(x - 3)$
Thus we had factorised the given expression.

For the equation ${x^2} - 9x + 18$, $x - 6$ and $x - 3$ are the factors.

Note: There is another method to solve the equation. It is called completing the square method.
For, the given equation can be rewritten as ${x^2} - 9x = - 18$
Now add the square of half of the coefficient of $x$ on both sides.
Here the coefficient of $x$ is $ - 9$. Half of $ - 9$ is $ - \dfrac{9}{2}$.
Squaring we get $\dfrac{{81}}{4}$.
Adding this on both sides we get,
${x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81}}{4} - 18$
$ \Rightarrow {x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81 - 72}}{4} = \dfrac{9}{4}$
We can observe that the left side is in the form ${a^2} - 2ab + {b^2}$ form which is equal to ${(a - b)^2}$.
So we get,
${x^2} - 9x + \dfrac{{81}}{4} = {(x - \dfrac{9}{2})^2}$
And also $\dfrac{9}{4} = {( \pm \dfrac{3}{2})^2}$.
So taking roots on both sides we can write,
$x - \dfrac{9}{2} = \pm \dfrac{3}{2}$
This gives $x = \dfrac{9}{2} + \dfrac{3}{2} = \dfrac{{12}}{2} = 6$ or $x = \dfrac{9}{2} - \dfrac{3}{2} = \dfrac{6}{2} = 3$
So the factors are $x - 6$ and $x - 3$.