
How do you divide $\dfrac{{{x}^{2}}+4x+4}{x+2}$ ?
Answer
524.1k+ views
Hint: In the given question, we are given a fraction which involves the polynomial in its numerator and denominator. Now we are asked to divide the polynomials present and hence we need to initially make the factors if possible, in numerator and denominator and then proceed.
Complete step-by-step solution:
Now, we are given a fraction in question which is $\dfrac{{{x}^{2}}+4x+4}{x+2}$. We are asked to divide the polynomials and then give the simplified answer. So, for this initially we will make factors of the numerator as
$\begin{align}
& {{x}^{2}}+4x+4 \\
& \Rightarrow {{x}^{2}}+2x+2x+4 \\
& \Rightarrow \left( {{x}^{2}}+2x \right)+\left( 2x+4 \right) \\
\end{align}$
Now, what we will do is we will find the common terms from the braces outside as $x\left( x+2 \right)+2\left( x+2 \right)$ . Now, further we can clearly see that we can take out the braces and hence it would be $\left( x+2 \right)\left( x+2 \right)$ .
So, now our given fraction becomes
$\begin{align}
& \dfrac{\left( x+2 \right)\left( x+2 \right)}{\left( x+2 \right)} \\
& \Rightarrow x+2 \\
\end{align}$
So, now we can see that the given fraction gets reduced to the above form in which we have $x+2$ term in common. Now, after cancelling the term$\left( x+2 \right)$from the numerator and denominator we get $x+2$.
Therefore, after dividing the given fraction $\dfrac{{{x}^{2}}+4x+4}{x+2}$from $x+2$ we get the answer as $x+2$only.
Note: The very common mistake we make in these types of questions is at the initial level we forget to make a factor in order to simplify the fraction and start dividing the given complex polynomials. Apart from this we try to match the coefficient rather than the power.
Complete step-by-step solution:
Now, we are given a fraction in question which is $\dfrac{{{x}^{2}}+4x+4}{x+2}$. We are asked to divide the polynomials and then give the simplified answer. So, for this initially we will make factors of the numerator as
$\begin{align}
& {{x}^{2}}+4x+4 \\
& \Rightarrow {{x}^{2}}+2x+2x+4 \\
& \Rightarrow \left( {{x}^{2}}+2x \right)+\left( 2x+4 \right) \\
\end{align}$
Now, what we will do is we will find the common terms from the braces outside as $x\left( x+2 \right)+2\left( x+2 \right)$ . Now, further we can clearly see that we can take out the braces and hence it would be $\left( x+2 \right)\left( x+2 \right)$ .
So, now our given fraction becomes
$\begin{align}
& \dfrac{\left( x+2 \right)\left( x+2 \right)}{\left( x+2 \right)} \\
& \Rightarrow x+2 \\
\end{align}$
So, now we can see that the given fraction gets reduced to the above form in which we have $x+2$ term in common. Now, after cancelling the term$\left( x+2 \right)$from the numerator and denominator we get $x+2$.
Therefore, after dividing the given fraction $\dfrac{{{x}^{2}}+4x+4}{x+2}$from $x+2$ we get the answer as $x+2$only.
Note: The very common mistake we make in these types of questions is at the initial level we forget to make a factor in order to simplify the fraction and start dividing the given complex polynomials. Apart from this we try to match the coefficient rather than the power.
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