Answer
385.2k+ views
Hint: Factorization is the process of writing a polynomial as a product of two or more terms. Usually these terms will leave a remainder of 0 when they divide the actual polynomial as they are the factors of it. We can factorize a given quadratic equation by two ways. Either by splitting the middle term or by directly applying the quadratic equation form and find its roots. So, we will continue with splitting the middle term process.
Complete step-by-step solution:
Now we have $f\left( x \right)=25{{x}^{2}}+10x+1$ .
Let us make use of a standard form of a quadratic equation to simplify things.
Let us factorize $f\left( x \right)$ by splitting the middle term.
Let us consider $h\left( x \right)=a{{x}^{2}}+bx+c$
We split the middle term, $bx$, in such a way so that we write it as the sum of two terms which when multiplied will give us the product of the first, $a{{x}^{2}}$, and last term,$c$, of the quadratic equation.
Upon comparing $f\left( x \right)$ with $h\left( x \right)$ , we conclude :
$\begin{align}
& \Rightarrow a=25 \\
& \Rightarrow b=10 \\
& \Rightarrow c=1 \\
\end{align}$
Now, let us split the middle term :
$\begin{align}
& \Rightarrow f\left( x \right)=25{{x}^{2}}+10x+1 \\
& \Rightarrow f\left( x \right)=25{{x}^{2}}+5x+5x+1 \\
\end{align}$
Since we know that $5x+5x=10x$ and $5x\times 5x=25{{x}^{2}}$
$\begin{align}
& \Rightarrow f\left( x \right)=25{{x}^{2}}+5x+5x+1 \\
& \Rightarrow f\left( x \right)=5x\left( 5x+1 \right)+1\left( 5x+1 \right) \\
\end{align}$
Taking $5x+1$ common , we get the following :
$\Rightarrow f\left( x \right)={{\left( 5x+1 \right)}^{2}}$
$\therefore $ We can factorize $f\left( x \right)=25{{x}^{2}}+10x+1$ as ${{\left( 5x+1 \right)}^{2}}$.
Note: We should be very careful while splitting the middle term. And comparing the equation and finding out the coefficients and plugging them in the formula must also be taken care of as they might lead to calculation errors and give wrong factors. If you find splitting the middle term to be difficult, please make use of the formula. If we are trying to factorize any polynomial with degree greater than 2 , we should try to guess at least one of it’s factors by trial and error method. Then divide it with the given polynomial to find all the factors.
Complete step-by-step solution:
Now we have $f\left( x \right)=25{{x}^{2}}+10x+1$ .
Let us make use of a standard form of a quadratic equation to simplify things.
Let us factorize $f\left( x \right)$ by splitting the middle term.
Let us consider $h\left( x \right)=a{{x}^{2}}+bx+c$
We split the middle term, $bx$, in such a way so that we write it as the sum of two terms which when multiplied will give us the product of the first, $a{{x}^{2}}$, and last term,$c$, of the quadratic equation.
Upon comparing $f\left( x \right)$ with $h\left( x \right)$ , we conclude :
$\begin{align}
& \Rightarrow a=25 \\
& \Rightarrow b=10 \\
& \Rightarrow c=1 \\
\end{align}$
Now, let us split the middle term :
$\begin{align}
& \Rightarrow f\left( x \right)=25{{x}^{2}}+10x+1 \\
& \Rightarrow f\left( x \right)=25{{x}^{2}}+5x+5x+1 \\
\end{align}$
Since we know that $5x+5x=10x$ and $5x\times 5x=25{{x}^{2}}$
$\begin{align}
& \Rightarrow f\left( x \right)=25{{x}^{2}}+5x+5x+1 \\
& \Rightarrow f\left( x \right)=5x\left( 5x+1 \right)+1\left( 5x+1 \right) \\
\end{align}$
Taking $5x+1$ common , we get the following :
$\Rightarrow f\left( x \right)={{\left( 5x+1 \right)}^{2}}$
$\therefore $ We can factorize $f\left( x \right)=25{{x}^{2}}+10x+1$ as ${{\left( 5x+1 \right)}^{2}}$.
Note: We should be very careful while splitting the middle term. And comparing the equation and finding out the coefficients and plugging them in the formula must also be taken care of as they might lead to calculation errors and give wrong factors. If you find splitting the middle term to be difficult, please make use of the formula. If we are trying to factorize any polynomial with degree greater than 2 , we should try to guess at least one of it’s factors by trial and error method. Then divide it with the given polynomial to find all the factors.
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