Question

How do you factor the expression ${x^2} - 15x + 54$?

Answer 1

First, equate this polynomial with zero and make it an equation. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
First, equate this polynomial with zero and make it an equation.
$\Rightarrow {x^2} - 15x + 54 = 0$
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now, compare ${x^2} - 15x + 54 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - 15x + 54 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 15$ and $c = 54$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 15} \right)^2} - 4\left( 1 \right)\left( {54} \right)$
After simplifying the result, we get
$\Rightarrow D = 225 - 216$
$\Rightarrow D = 9$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$\Rightarrow x = \dfrac{{ - \left( { - 15} \right) \pm 3}}{{2 \times 1}}$
It can be further simplified as
$\Rightarrow x = \dfrac{{15 \pm 3}}{2}$
$\Rightarrow x = \dfrac{{15 + 3}}{2}$ and $x = \dfrac{{15 - 3}}{2}$
$\Rightarrow x = \dfrac{{18}}{2}$ and $x = \dfrac{{12}}{2}$
$\Rightarrow x = 9$ and $x = 6$
$\Rightarrow x - 9 = 0$ and $x - 6 = 0$

Final Solution: Therefore, the trinomial ${x^2} - 15x + 54$ can be factored as $\left( {x - 6} \right)\left( {x - 9} \right)$.

Note: We can also factorize a given trinomial by splitting the middle term.
For factorising an algebraic expression of the type $a{x^2} + bx + c$, we find two factors $p$ and $q$ such that
$ac = pq$ and $p + q = b$
Step by step solution:
Given, ${x^2} - 15x + 54$
We have to factor this trinomial.
To factor this trinomial, first we have to find the product of the first and last constant term of the expression.
Here, the first constant term in ${x^2} - 15x + 54$ is $1$, as it is the coefficient of ${x^2}$ and last constant term is $54$, as it is a constant value.
Now, we have to multiply the coefficient of ${x^2}$ with the constant value in ${x^2} - 15x + 54$, i.e., multiply $1$ with $54$.
Multiplying $1$ and $54$, we get
$1 \times 54 = 54$
Now, we have to find the factors of $54$ in such a way that addition or subtraction of those factors is the middle constant term.
Middle constant term or coefficient of $x$ in ${x^2} - 15x + 54$ is $- 15$.
So, we have to find two factors of $54$, which on multiplying gives $54$ and in addition gives $- 15$.
We can do this by determining all factors of $54$.
Factors of $54$ are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 9$.
Now among these values find two factors of $54$, which on multiplying gives $54$ and in addition gives $- 15$.
After observing, we can see that
$\left( { - 6} \right) \times \left( { - 9} \right) = 54$ and $\left( { - 6} \right) + \left( { - 9} \right) = - 15$
So, these factors are suitable for factorising the given trinomial.
Now, the next step is to split the middle constant term or coefficient of $x$ in these factors.
That is, write $- 15x$ as $- 6x - 9x$ in ${x^2} - 15x + 54$.
After writing$- 15x$ as $- 6x - 9x$ in ${x^2} - 15x + 54$, we get
$\Rightarrow {x^2} - 15x + 54 = {x^2} - 6x - 9x + 54$
Now, taking $x$ common in ${x^2} - 6x$ and putting in above equation, we get
$\Rightarrow {x^2} - 15x + 54 = x\left( {x - 6} \right) - 9x + 54$
Now, taking $\left( { - 9} \right)$ common in $- 9x + 54$ and putting in above equation, we get
$\Rightarrow {x^2} - 15x + 54 = x\left( {x - 6} \right) - 9\left( {x - 6} \right)$
Now, taking $\left( {x - 6} \right)$ common in $x\left( {x - 6} \right) - 9\left( {x - 6} \right)$ and putting in above equation, we get
$\Rightarrow {x^2} - 15x + 54 = \left( {x - 6} \right)\left( {x - 9} \right)$
Final Solution: Therefore, the trinomial ${x^2} - 15x + 54$ can be factored as $\left( {x - 6} \right)\left( {x - 9} \right)$.
In the above question, it should be noted that we took $- 6$ and $- 9$ as factors of $54$, which on multiplying gives $54$ and in addition gives $- 15$. No, other factors will satisfy the condition. If we take wrong factors, then we will not be able to take common terms out in the next step. So, carefully select the numbers.