
How do you simply $\left( x-5 \right)\left( x+3 \right)$?
Answer
543.3k+ views
Hint: For these kinds of questions, all we need to use is simple mathematics. Before starting any kind of chapter, we all have learnt about a few laws such as commutative law, associative law and distributive law. Now , we are going to make use of distributive law. Distributive law states the following :
$a\left( b+c \right)=a\times b+a\times c$ where $a,b,c$ can be numbers or variables as this law holds good even in algebra.
Complete step by step solution:
In the given question, we have $\left( x-5 \right)\left( x+3 \right)$. So we just have to distribute or in mathematical terms, we have to multiply one bracket terms with the other.
Multiplying the terms of one bracket with the terms of another bracket is nothing but exercising distributive law.
Distributive law states $a\left( b+c \right)=a\times b+a\times c$.
So let us multiply the term by applying distributive law.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow \left( x-5 \right)\left( x+3 \right) \\
& \Rightarrow x\left( x+3 \right)-5\left( x+3 \right) \\
& \Rightarrow {{x}^{2}}+3x-5x-15 \\
\end{align}$
Now let us gather all the $x$ terms together .
Upon simplifying further, we get the following :
$\begin{align}
& \Rightarrow \left( x-5 \right)\left( x+3 \right) \\
& \Rightarrow x\left( x+3 \right)-5\left( x+3 \right) \\
& \Rightarrow {{x}^{2}}+3x-5x-15 \\
& \Rightarrow {{x}^{2}}-2x-15 \\
\end{align}$
So we got a quadratic equation at the end. So the question with which we started is nothing by the factors of the quadratic equation that we obtained at the end i.e ${{x}^{2}}-2x-15$.
$\therefore $ Hence, upon solving $\left( x-5 \right)\left( x+3 \right)$, we get ${{x}^{2}}-2x-15$.
Note: In the answer, $\left( x+3 \right)$ is being distributed to $\left( x-5 \right)$ but the reverse can also happen. So $\left( x-5 \right)$ can also be distributed to $\left( x+3 \right)$ and we would still end up with the same result. All these laws hold good for algebra as these laws are already proved for them but as we move to different topics and fields in mathematics such as complex numbers, we should check whether these laws hold good in that field of mathematics before using them.
$a\left( b+c \right)=a\times b+a\times c$ where $a,b,c$ can be numbers or variables as this law holds good even in algebra.
Complete step by step solution:
In the given question, we have $\left( x-5 \right)\left( x+3 \right)$. So we just have to distribute or in mathematical terms, we have to multiply one bracket terms with the other.
Multiplying the terms of one bracket with the terms of another bracket is nothing but exercising distributive law.
Distributive law states $a\left( b+c \right)=a\times b+a\times c$.
So let us multiply the term by applying distributive law.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow \left( x-5 \right)\left( x+3 \right) \\
& \Rightarrow x\left( x+3 \right)-5\left( x+3 \right) \\
& \Rightarrow {{x}^{2}}+3x-5x-15 \\
\end{align}$
Now let us gather all the $x$ terms together .
Upon simplifying further, we get the following :
$\begin{align}
& \Rightarrow \left( x-5 \right)\left( x+3 \right) \\
& \Rightarrow x\left( x+3 \right)-5\left( x+3 \right) \\
& \Rightarrow {{x}^{2}}+3x-5x-15 \\
& \Rightarrow {{x}^{2}}-2x-15 \\
\end{align}$
So we got a quadratic equation at the end. So the question with which we started is nothing by the factors of the quadratic equation that we obtained at the end i.e ${{x}^{2}}-2x-15$.
$\therefore $ Hence, upon solving $\left( x-5 \right)\left( x+3 \right)$, we get ${{x}^{2}}-2x-15$.
Note: In the answer, $\left( x+3 \right)$ is being distributed to $\left( x-5 \right)$ but the reverse can also happen. So $\left( x-5 \right)$ can also be distributed to $\left( x+3 \right)$ and we would still end up with the same result. All these laws hold good for algebra as these laws are already proved for them but as we move to different topics and fields in mathematics such as complex numbers, we should check whether these laws hold good in that field of mathematics before using them.
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