
Factorise \[{x^2} - 16\].
Answer
523.2k+ views
Hint: Factoring is the process in which an expression or number is expressed as the product of its factors, that is by factorising we obtain numbers or terms which can be multiplied to get back the original expression. In order to factorise the given expression, express it as a difference of two squares. Then use the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].
Complete step by step solution:
To factorise the given expression, we have to break it into terms which can be multiplied to get back the original expression.
To do so first express the given equation as a difference of two squares.
Given,
\[{x^2} - 16\]
\[ = {x^2} - {4^2}\]
Now use the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
\[\left( {x + 4} \right)\left( {x - 4} \right)\]
Hence , \[{x^2} - 16 = \left( {x + 4} \right)\left( {x - 4} \right)\].
Thus, the factors of \[{x^2} - 16\] are \[\left( {x + 4} \right)\] and \[\left( {x - 4} \right)\].
Additional Information: The above expression if equated with \[0\], becomes a quadratic equations, and the solutions can be found by using the formula,
if \[\left( {a + b} \right)\left( {a - b} \right) = 0\], then either \[\left( {a + b} \right) = 0\] or \[\left( {a - b} \right) = 0\].
That is if,
\[{x^2} - 16 = 0\]
\[ \Rightarrow \left( {x + 4} \right)\left( {x - 4} \right) = 0\]
\[ \Rightarrow \left( {x + 4} \right) = 0\] or \[\left( {x - 4} \right) = 0\]
\[ \Rightarrow x = \pm 4\].
Note: Factorization means only expressing the given term as a product of its factors, and not finding the solutions. Also note that the solution of any quadratic equation of the form, \[a{x^2} + bx + c = 0\], can be solved using the Sridharacharya method which states,\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Complete step by step solution:
To factorise the given expression, we have to break it into terms which can be multiplied to get back the original expression.
To do so first express the given equation as a difference of two squares.
Given,
\[{x^2} - 16\]
\[ = {x^2} - {4^2}\]
Now use the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
\[\left( {x + 4} \right)\left( {x - 4} \right)\]
Hence , \[{x^2} - 16 = \left( {x + 4} \right)\left( {x - 4} \right)\].
Thus, the factors of \[{x^2} - 16\] are \[\left( {x + 4} \right)\] and \[\left( {x - 4} \right)\].
Additional Information: The above expression if equated with \[0\], becomes a quadratic equations, and the solutions can be found by using the formula,
if \[\left( {a + b} \right)\left( {a - b} \right) = 0\], then either \[\left( {a + b} \right) = 0\] or \[\left( {a - b} \right) = 0\].
That is if,
\[{x^2} - 16 = 0\]
\[ \Rightarrow \left( {x + 4} \right)\left( {x - 4} \right) = 0\]
\[ \Rightarrow \left( {x + 4} \right) = 0\] or \[\left( {x - 4} \right) = 0\]
\[ \Rightarrow x = \pm 4\].
Note: Factorization means only expressing the given term as a product of its factors, and not finding the solutions. Also note that the solution of any quadratic equation of the form, \[a{x^2} + bx + c = 0\], can be solved using the Sridharacharya method which states,\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
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