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Evaluate: \[(c + 5)(c - 3)\].

Answer
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261k+ 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).
Last updated date: 26th Sep 2023
Total views: 261k
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