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# How do you factor $9{{x}^{2}}-25$ ?

Last updated date: 13th Jun 2024
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Hint:
We first try to explain the concept of factorisation and the ways a factorisation of a polynomial can be done. We use the identity theorem of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor the given polynomial $9{{x}^{2}}-25$ . We assume the values of $a=3x;b=5$ . The final multiplied linear polynomials are the solution of the problem.

For the factorisation of the given quadratic polynomial $9{{x}^{2}}-25$ , we apply the factorisation identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .
We get $9{{x}^{2}}-25={{\left( 3x \right)}^{2}}-{{5}^{2}}$ . We put the value of $a=3x;b=5$ .
$9{{x}^{2}}-25={{\left( 3x \right)}^{2}}-{{5}^{2}}=\left( 3x+5 \right)\left( 3x-5 \right)$ .
Therefore, the final factorisation of $9{{x}^{2}}-25$ is $\left( 3x+5 \right)\left( 3x-5 \right)$ .
The formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ derives from the solution identity of
\begin{align} & {{a}^{2}}-{{b}^{2}} \\ & ={{a}^{2}}-ab+ab-{{b}^{2}} \\ & =a\left( a-b \right)+b\left( a-b \right) \\ & =\left( a+b \right)\left( a-b \right) \\ \end{align}