
How do you factor the expression $9{{x}^{2}}-1$ ?
Answer
540.3k+ views
Hint:To factorize the expression $9{{x}^{2}}-1$ , we will use algebraic identities. We can rewrite the polynomial with each term in square as ${{\left( 3x \right)}^{2}}-1$ . Now, this is in the form of the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . By substituting for a and b, we will get the required answer.
Complete step by step answer:
We need to factorize $9{{x}^{2}}-1$ . The given question is a polynomial. We can define a polynomial as an expression which is composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, multiplication and division.
To factorize a polynomial, we commonly use the algebraic identities. First, let us rewrite the given expression. $9{{x}^{2}}$ can be written as $3\times 3\times x\times x={{\left( 3x \right)}^{2}}$
$\begin{align}
& 9{{x}^{2}}-1 \\
& ={{\left( 3x \right)}^{2}}-1 \\
\end{align}$
We can write the second terms 1 as ${{1}^{2}}$ . Therefore, the above expression becomes
${{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}$ .
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .
Let us compare the identity with ${{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}$ . We can see that $a=3x$ and $b=1$ . Hence, we can write the given expression as:
$\Rightarrow {{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}=\left( 3x+1 \right)\left( 3x-1 \right)$
Hence, the $9{{x}^{2}}-1$ can be factorized to $\left( 3x+1 \right)\left( 3x-1 \right)$ .
Note:
Whenever the factors of a polynomial are to be found, students must keep in mind to convert or rewrite the given question in the form of algebraic identities. Algebraic identities are algebraic equations which are always true for every value of variables in them. The most commonly used algebraic identities are ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\text{ and }{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . There are also other identities for polynomials of order 3.
Complete step by step answer:
We need to factorize $9{{x}^{2}}-1$ . The given question is a polynomial. We can define a polynomial as an expression which is composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, multiplication and division.
To factorize a polynomial, we commonly use the algebraic identities. First, let us rewrite the given expression. $9{{x}^{2}}$ can be written as $3\times 3\times x\times x={{\left( 3x \right)}^{2}}$
$\begin{align}
& 9{{x}^{2}}-1 \\
& ={{\left( 3x \right)}^{2}}-1 \\
\end{align}$
We can write the second terms 1 as ${{1}^{2}}$ . Therefore, the above expression becomes
${{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}$ .
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .
Let us compare the identity with ${{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}$ . We can see that $a=3x$ and $b=1$ . Hence, we can write the given expression as:
$\Rightarrow {{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}=\left( 3x+1 \right)\left( 3x-1 \right)$
Hence, the $9{{x}^{2}}-1$ can be factorized to $\left( 3x+1 \right)\left( 3x-1 \right)$ .
Note:
Whenever the factors of a polynomial are to be found, students must keep in mind to convert or rewrite the given question in the form of algebraic identities. Algebraic identities are algebraic equations which are always true for every value of variables in them. The most commonly used algebraic identities are ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\text{ and }{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . There are also other identities for polynomials of order 3.
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