Question

How do you factor and solve $16{{x}^{2}}-40x+25=0$ ?

Answer 1

Hint: So in the question, we are asked to factorize and solve for the value of $x$ . In general, 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. So, we will use splitting the middle term method.

Complete step by step answer:
Now we have $f\left( x \right)=16{{x}^{2}}-40x+25=0$ .
Let’s make use of a standard form of a quadratic equation to simplify things.
Let’s factorize $f\left( x \right)$ by splitting the middle term.
Let $h\left( x \right)=a{{x}^{2}}+bx+c$
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 :
$\begin{align}
& \Rightarrow a=16 \\
& \Rightarrow b=-40 \\
& \Rightarrow c=25 \\
\end{align}$
Now , let’s split the middle term :
$\begin{align}
& \Rightarrow f\left( x \right)=16{{x}^{2}}-40x+25=0 \\
& \Rightarrow f\left( x \right)=16{{x}^{2}}-20x-20x+25=0 \\
\end{align}$
Since $-20x-20x=-40x$ and $-20x\times -20x=400{{x}^{2}}=16{{x}^{2}}\times 25$
$\begin{align}
& \Rightarrow f\left( x \right)=16{{x}^{2}}-20x-20x+25=0 \\
& \Rightarrow f\left( x \right)=4x\left( 4x-5 \right)-5\left( 4x-5 \right)=0 \\
\end{align}$
Taking $4x-5$ common. Upon doing so, we get the following :
$\Rightarrow f\left( x \right)={{\left( 4x-5 \right)}^{2}}=0$
Upon factoring \[f\left( x \right)=16{{x}^{2}}-40x+25=0\],
We got $f\left( x \right)={{\left( 4x-5 \right)}^{2}}=0$ .
Now let us solve for $x$ .
$\Rightarrow {{\left( 4x-5 \right)}^{2}}=0$
Square rooting on both sides. Upon doing so, we get the following :
$\Rightarrow \left( 4x-5 \right)=0$ .
 Now let us send the -5 onto the right hand side of the equation. Upon doing so, we get the following :
$\Rightarrow 4x=5$ .
Now let us divide the entire equation by 4. Upon doing so, we get the following :
$\Rightarrow x=\dfrac{5}{4}$ .
Now we might wonder why we get only one value of $x$ when we are supposed to get 2 values as it is a quadratic equation.
The answer to the question is that $x=\dfrac{5}{4}$ is a repeated root. Both the roots of the quadratic equation \[f\left( x \right)=16{{x}^{2}}-40x+25=0\] are $x=\dfrac{5}{4}$ .

$\therefore $ Hence after factoring and solving the equation \[16{{x}^{2}}-40x+25=0\] we get the value of $x$ as $\dfrac{5}{4}$.

Note: We should not worry when we are getting only one value of $x$ . And in times when we do get , we should check the discriminant. Plug – in the value of all coefficients in $\sqrt{{{b}^{2}}-4ac}$ and if we obtain a 0, that means that particular quadratic equation is having repeated roots. This is one way to cross check our answer. Should always be careful while splitting the middle term. And if we find this method difficult, we can always go for the quadratic equation formula. Just plug – in the values and we get our answers.