Question

How do you factor the expression ${x^2} - 7x - 18$?

Answer 1

Hint: We will use the method of splitting the middle term to factorize the given polynomial. In order to do this, we will write the middle term as a sum of two terms. Then, we will club together like terms and take common factors out. Finally, we will get the polynomial as a product of two factors.

Complete step by step solution:
The quadratic polynomial given to us is ${x^2} - 7x - 18$.
We have to factorize this polynomial i.e. we have to find two factors such that the given polynomial can be expressed as a product of the two factors. To do this, we will use the method of splitting the middle term.
In the given polynomial ${x^2} - 7x - 18$, we have to split $ - 7$ as a sum of two terms whose product is $1 \times ( - 18) = - 18$.
Let us find the factors of 18 and find combinations of numbers that on addition or subtraction will give us $ - 7$.
We know that $18 = 2 \times 3 \times 3$.
We can also write this as $18 = 2 \times 9$.
We also know that $ - 7 = ( - 9) + 2$.
 So, the required numbers are $ - 9$ and $2$. Hence, the polynomial can be written as
${x^2} - 7x - 18 = {x^2} - 9x + 2x - 18$
We will club the first two terms and the last two terms together. In the first two terms the common factor is $x$. In the last two terms, the common factor is $2$. Thus,
$ \Rightarrow {x^2} - 7x - 18 = x(x - 9) + 2(x - 9)$
We see on the RHS that the factor $(x - 9)$ is common to both terms. Factoring out the common terms, we get

$ \Rightarrow {x^2} - 7x - 18 = (x - 9)(x + 2)$

Note:
To factorize a polynomial $a{x^2} + bx + c$ by splitting the middle term, we have to find two terms such that we can write \[b\] as a sum of the two terms such that their product is $a \times c$. This means that we find two numbers $p$ and $q$ such that $p + q = b$ and $pq = ac$. After finding $p$ and $q$, we split the middle term in the quadratic polynomial as $px + qx$ and get the desired factors by grouping the terms. Here, the terms $p$ and $q$ are not necessarily a positive term.