Answer
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Hint: In the question we need to factorise the given equation. We can observe that the given equation is a quadratic equation. Generally, we will follow different methods for different types of equations. For the quadratic equation which is in the form of $a{{x}^{2}}+bx+c$, we will break the middle term into two parts like $b={{x}_{1}}+{{x}_{2}}$ such that ${{x}_{1}}.{{x}_{2}}=ac$. So, we will first calculate the value of $ac$ and we will write the factors of the calculated $ac$ and choose two factors such that they satisfy the above-mentioned conditions. After getting the values of ${{x}_{1}}$, ${{x}_{2}}$ we will break the middle term and simplify the equation by taking appropriate terms as common. After taking the terms as common we will get our required result.
Given that, ${{x}^{2}}-8x+16$.
Comparing the above equation with $a{{x}^{2}}+bx+c$, then we can write
Coefficient of ${{x}^{2}}$ is $a=1$.
Coefficient of $x$ is $b=-8$.
Constant is $c=16$.
Now the value of $ac$ is given by
$\begin{align}
& ac=1\left( 16 \right) \\
& \Rightarrow ac=16 \\
\end{align}$
Here we have the value of $ac$ as $16$. We know that the factors for $16$ are $1$, $2$, $4$, $8$, $16$, $45$.
From the above factors we can observe that
$\begin{align}
& -4-4=8 \\
& -4\times -4=16 \\
\end{align}$
So, we can break the middle term $-8x$ as $-8x=-4x-4x$. From this value, we are going to write the given equation as
$\Rightarrow {{x}^{2}}-8x+16={{x}^{2}}-4x-4x+16$
Taking $x$ as common from ${{x}^{2}}-4x$ and $-4$ as common from $-4x+16$, then the above equation is modified as
$\Rightarrow {{x}^{2}}-8x+16=x\left( x-4 \right)-4\left( x-4 \right)$
Again, taking $x-4$ as common on RHS of the above equation, then we will get
$\Rightarrow {{x}^{2}}-8x+16=\left( x-4 \right)\left( x-4 \right)$
Hence the factors of the given equation are $x-4$ and $x-4$.
We can also write the above equation as
$\Rightarrow {{x}^{2}}-8x+16={{\left( x-4 \right)}^{2}}$
Note:
In this problem, they have only asked to factorize the given equation, so we have ended our procedure after calculating the factors. If they have asked to solve the equation or asked to find the roots, then we need to equate the given equation to zero and we will find the values of $x$.
Given that, ${{x}^{2}}-8x+16$.
Comparing the above equation with $a{{x}^{2}}+bx+c$, then we can write
Coefficient of ${{x}^{2}}$ is $a=1$.
Coefficient of $x$ is $b=-8$.
Constant is $c=16$.
Now the value of $ac$ is given by
$\begin{align}
& ac=1\left( 16 \right) \\
& \Rightarrow ac=16 \\
\end{align}$
Here we have the value of $ac$ as $16$. We know that the factors for $16$ are $1$, $2$, $4$, $8$, $16$, $45$.
From the above factors we can observe that
$\begin{align}
& -4-4=8 \\
& -4\times -4=16 \\
\end{align}$
So, we can break the middle term $-8x$ as $-8x=-4x-4x$. From this value, we are going to write the given equation as
$\Rightarrow {{x}^{2}}-8x+16={{x}^{2}}-4x-4x+16$
Taking $x$ as common from ${{x}^{2}}-4x$ and $-4$ as common from $-4x+16$, then the above equation is modified as
$\Rightarrow {{x}^{2}}-8x+16=x\left( x-4 \right)-4\left( x-4 \right)$
Again, taking $x-4$ as common on RHS of the above equation, then we will get
$\Rightarrow {{x}^{2}}-8x+16=\left( x-4 \right)\left( x-4 \right)$
Hence the factors of the given equation are $x-4$ and $x-4$.
We can also write the above equation as
$\Rightarrow {{x}^{2}}-8x+16={{\left( x-4 \right)}^{2}}$
Note:
In this problem, they have only asked to factorize the given equation, so we have ended our procedure after calculating the factors. If they have asked to solve the equation or asked to find the roots, then we need to equate the given equation to zero and we will find the values of $x$.
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