Question

How do you factor ${x^2} + 5x + 24$?

Answer 1

Hint: The given equation in the question is a quadratic equation. It is an equation containing a single variable of degree $2$. Its general form is $a{x^2} + bx + c = 0$, where $x$ is the variable and $a$, $b$, $c$ are the constants. Also, the degree is the highest power (in the whole number) that a variable has, in an equation. There are two methods to factorize a quadratic equation given the following:
3. In the first method, to factorise a quadratic equation $a{x^2} + bx + c = 0$, we will have to find out two numbers that multiply to give $ac$ and add to give $b$. If the two numbers are $p$ and $q$, then we get the factors as $(x + p)$ and $(x + q)$ such that $a{x^2} + bx + c = (x + p)(x + q)$.
4. In the second method, to factorise a quadratic equation $a{x^2} + bx + c = 0$, we will find two numbers $p$ and $q$ which will be
$p = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $q = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$, such that $a{x^2} + bx + c = (x + p)(x + q)$.

Complete step by step solution:
The given quadratic expression is ${x^2} + 5x + 24$. If we want to find the factors, we will have to convert it into an equation by this way:
${x^2} + 5x + 24 = 0$
Here $a = 1$; $b = 5$ and $c = 24$.
We will at first use the first method. So in order to factorize the quadratic equation, we will have to find out two numbers that multiply to give $ac$, i.e. $24$ here, and add to give $b$, i.e. $5$ here. Now for this, we will find all the factors of $24$, such that
$ \Rightarrow 24 = 2 \times 2 \times 2 \times 3$
Now we will multiply all these four factors of $24$ within themselves and arrange them in such a manner that when we add or subtract them, we must get $5$ as the answer and when we multiply them, we must get $24$ as the answer.
When we try to do so, we find no such two numbers to satisfy the given conditions. Therefore we will now use the second method.
Since we have $a = 1$; $b = 5$ and $c = 24$, so substituting these values for $p$, we shall get
$ \Rightarrow p = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow p = \dfrac{{ - 5 + \sqrt {{5^2} - 4 \times 1 \times 24} }}{{2 \times 1}}$
$ \Rightarrow p = \dfrac{{ - 5 + \sqrt {25 - 96} }}{2}$
$ \Rightarrow p = \dfrac{{ - 5 + \sqrt { - 71} }}{2}$
Similarly, for $q$, we shall get
$ \Rightarrow q = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow q = \dfrac{{ - 5 - \sqrt {{5^2} - 4 \times 1 \times 24} }}{{2 \times 1}}$
$ \Rightarrow q = \dfrac{{ - 5 - \sqrt {25 - 96} }}{2}$
$ \Rightarrow q = \dfrac{{ - 5 - \sqrt { - 71} }}{2}$
Hence, we have $p = \dfrac{{ - 5 + \sqrt { - 71} }}{2}$ and $q = \dfrac{{ - 5 - \sqrt { - 71} }}{2}$. So the factors of ${x^2} + 5x + 24$ become $[x - (\dfrac{{ - 5 + \sqrt { - 71} }}{2})]$ and $[x - (\dfrac{{ - 5 - \sqrt { - 71} }}{2})]$, such that

${x^2} + 5x + 24 = [x - (\dfrac{{ - 5 + \sqrt { - 71} }}{2})][x - (\dfrac{{ - 5 - \sqrt { - 71} }}{2})]$.

Note:
In the given answer, $\sqrt { - 71} $ is an imaginary number which can also be written as $71i$. Here $i$ is known as iota such that $i = \sqrt { - 1} $. Hence we can also write the factors as
${x^2} + 5x + 24 = [x - (\dfrac{{ - 5 + 71i}}{2})][x - (\dfrac{{ - 5 - 71i}}{2})]$.