Factorise \[{x^2} - 16\].
Answer
560.1k+ 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,\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Recently Updated Pages
Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Trending doubts
Find the sum of series 1 + 2 + 3 + 4 + 5 + + 100 class 9 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

What is the Full Form of ICSE / ISC ?

What is pollution? How many types of pollution? Define it

Name 10 Living and Non living things class 9 biology CBSE

