
Evaluate: \[(c + 5)(c - 3)\].
Answer
519.9k+ views
Hint: It is a simple algebraic expression. we can solve this either by directly using the algebraic formula \[(a + b)(a - c) = {a^2} + a(b - c) - bc\] to obtain the final expression or we can simplify the given expression by multiplying the terms to get the required expression.
Complete step by step answer:
Given: \[(c + 5)(c - 3)\]
It can be written as
\[c(c - 3) + 5(c - 3)\]
By further calculation
\[{c^2} - 3c + 5c - 15\]
on simplification
\[{c^2} + 2c - 15\]
Further if you simplify the obtained expression by taking the common factor, we will get back the given expression that the obtained solution is \[{c^2} + 2c - 15\]. Simplifying the above using factorization method we get factors as \[5\] and \[ - 3\] with this above expression can be written as \[{c^2} + 5c - 3c - 15\]. Taking common factor and rearranging
\[c(c + 5) - 3(c + 5)\]
Again, taking common factor we get
\[(c + 5)(c - 3)\]
Note: the given expression is of the form \[(a + b)(a - b)\]so we can directly use the formula
\[(a + b)(a - c) = {a^2} + a(b - c) - bc\]
Where \[a = c,b = 5,c = 3\]
on substitution in the above formula we get,
\[(c + 5)(c - 3) = {c^2} + c(5 - 3) - (5)(3)\]
\[\Rightarrow {c^2} + 5c - 3c - 15\]
\[\Rightarrow c(c + 5) - 3(c + 5)\]
Further if you simplify the obtained expression by taking the common factor, we will get back the given expression that the obtained solution is \[{c^2} + 2c - 15\]. Simplifying the above using factorization method we get factors as \[5\] and \[ - 3\] with this above expression can be written as \[{c^2} + 5c - 3c - 15\]. (taking common factor and rearranging)
\[ c(c + 5) - 3(c + 5)\]
Again, taking common factor we get,
\[(c + 5)(c - 3)\]
The solution is the same in both cases. In the above problem we cannot simplify the equation further to get the value of c because the given expression is not equated to zero. (that is supposed if the given equation is of the form \[(c + 5)(c - 3) = 0\] then we can simplify it by using Factorization method or formula method).
Complete step by step answer:
Given: \[(c + 5)(c - 3)\]
It can be written as
\[c(c - 3) + 5(c - 3)\]
By further calculation
\[{c^2} - 3c + 5c - 15\]
on simplification
\[{c^2} + 2c - 15\]
Further if you simplify the obtained expression by taking the common factor, we will get back the given expression that the obtained solution is \[{c^2} + 2c - 15\]. Simplifying the above using factorization method we get factors as \[5\] and \[ - 3\] with this above expression can be written as \[{c^2} + 5c - 3c - 15\]. Taking common factor and rearranging
\[c(c + 5) - 3(c + 5)\]
Again, taking common factor we get
\[(c + 5)(c - 3)\]
Note: the given expression is of the form \[(a + b)(a - b)\]so we can directly use the formula
\[(a + b)(a - c) = {a^2} + a(b - c) - bc\]
Where \[a = c,b = 5,c = 3\]
on substitution in the above formula we get,
\[(c + 5)(c - 3) = {c^2} + c(5 - 3) - (5)(3)\]
\[\Rightarrow {c^2} + 5c - 3c - 15\]
\[\Rightarrow c(c + 5) - 3(c + 5)\]
Further if you simplify the obtained expression by taking the common factor, we will get back the given expression that the obtained solution is \[{c^2} + 2c - 15\]. Simplifying the above using factorization method we get factors as \[5\] and \[ - 3\] with this above expression can be written as \[{c^2} + 5c - 3c - 15\]. (taking common factor and rearranging)
\[ c(c + 5) - 3(c + 5)\]
Again, taking common factor we get,
\[(c + 5)(c - 3)\]
The solution is the same in both cases. In the above problem we cannot simplify the equation further to get the value of c because the given expression is not equated to zero. (that is supposed if the given equation is of the form \[(c + 5)(c - 3) = 0\] then we can simplify it by using Factorization method or formula method).
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest States of India?

What is the theme or message of the poem The road not class 9 english CBSE

Define development

Distinguish between population growth and population class 9 social science CBSE

Explain the importance of pH in everyday life class 9 chemistry CBSE

The winter rain in Chennai is caused by A SouthWest class 9 social science CBSE


