
How do you factor the expression ${{x}^{2}}-169$?
Answer
560.4k+ views
Hint: Here in this question, we have to solve the expression by using algebraic identities. We will use algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factorise the given expression. When we will multiply a number two times, it means we are squaring that number. Similarly if we square 13 two times i.e. $13\times 13$ it will be ${{13}^{2}}$ which is equal to 169.
Complete step by step answer:
Now, let’s solve the question.
If any number or variable is multiplied with itself a particular number of times, it defines it’s power. For example: If we multiply a variable ‘x’ 5 times, it will be $x\times x\times x\times x\times x$ which can be written as ${{x}^{5}}$. In the same way if we multiply 4 three times, it will be $4\times 4\times 4$ which can be written as ${{4}^{3}}=64$. And if the power is 2 that means that particular number or variable is multiplied two times. In the same way, if we see the question, here 169 is formed by multiplying 13 two times. So it can be seen like this:
$\Rightarrow {{x}^{2}}-{{13}^{2}}......(i)$
As we know some of the algebraic identities. Let’s discuss them.
$\begin{align}
& \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
& \Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
& \Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
& \Rightarrow \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab \\
\end{align}$
As we can see that ${{x}^{2}}-{{13}^{2}}$is in the form of ${{a}^{2}}-{{b}^{2}}$.
So by applying, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in equation(i) we get:
$\Rightarrow \left( x+13 \right)\left( x-13 \right)$
These are our final factors of the expression ${{x}^{2}}-169$.
Note: Do remember all the identities of algebraic expressions. Students should know the square and square roots of the numbers at least 1 to 20. This will be helpful in factoring the expression. By looking at the question, you should be able to identify which identity will get fit for that particular expression.
Complete step by step answer:
Now, let’s solve the question.
If any number or variable is multiplied with itself a particular number of times, it defines it’s power. For example: If we multiply a variable ‘x’ 5 times, it will be $x\times x\times x\times x\times x$ which can be written as ${{x}^{5}}$. In the same way if we multiply 4 three times, it will be $4\times 4\times 4$ which can be written as ${{4}^{3}}=64$. And if the power is 2 that means that particular number or variable is multiplied two times. In the same way, if we see the question, here 169 is formed by multiplying 13 two times. So it can be seen like this:
$\Rightarrow {{x}^{2}}-{{13}^{2}}......(i)$
As we know some of the algebraic identities. Let’s discuss them.
$\begin{align}
& \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
& \Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
& \Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
& \Rightarrow \left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab \\
\end{align}$
As we can see that ${{x}^{2}}-{{13}^{2}}$is in the form of ${{a}^{2}}-{{b}^{2}}$.
So by applying, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in equation(i) we get:
$\Rightarrow \left( x+13 \right)\left( x-13 \right)$
These are our final factors of the expression ${{x}^{2}}-169$.
Note: Do remember all the identities of algebraic expressions. Students should know the square and square roots of the numbers at least 1 to 20. This will be helpful in factoring the expression. By looking at the question, you should be able to identify which identity will get fit for that particular expression.
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