Factorise \[{x^2} - 16\].
Answer
553.5k+ 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 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Social Science: Engaging Questions & Answers for Success

Trending doubts
21st March is celebrated as A World environment day class 9 biology CBSE

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

What is the Full Form of ISI and RAW

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

Difference Between Plant Cell and Animal Cell

Here 1 km is equal to how many miles class 9 maths CBSE

