Question

How do you factor \[5{{x}^{2}}-26x+5\]?

Answer 1

Hint: From the question given, we have been asked to factor the quadratic expression \[5{{x}^{2}}-26x+5\]. We can find the factors for the given quadratic expression by using the process of factorization. By using the process of factorization, we can find the factors for the given quadratic expression. First of all, we have to know about the process of factorization. By using the product of coefficients and writing it as the sum of the two numbers we will find out the factors. This is the process of factorization.

Complete step by step answer:
From the question given, we have been a quadratic expression \[5{{x}^{2}}-26x+5\]
First of all, we have to multiply the coefficient of \[{{x}^{2}}\] and constant.
By multiplying the coefficient of \[{{x}^{2}}\] and constant, we get \[\Rightarrow 5\times 5=25\]
Now, we have to write the coefficient of \[x\] as the sum of two numbers and the numbers should be the factors of \[25\].
\[\Rightarrow 25=-1\times -25\]
By writing the given expression as above said, we get \[\Rightarrow 5{{x}^{2}}-25x-x+5\]
Now, we have to take the common terms out and simplify further to obtain the factors.
By taking the common terms out and simplifying further, we get
\[\Rightarrow 5x\left( x-5 \right)-1\left( x-5 \right)\]
\[\Rightarrow \left( 5x-1 \right)\left( x-5 \right)\]
Therefore, \[5{{x}^{2}}-26x+5=\left( 5x-1 \right)\left( x-5 \right)\]
Hence, we got the factors for the given quadratic expression by using the process of factorization.

Note:
We should be well aware of the process of factorization. Also, we should be very careful while writing the middle term as a sum of two numbers which is equal to the product of coefficient of first term and constant. Also, we should be very careful while simplifying the expression. This question can also be answered by using the formulae for finding the roots of any quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . For $5{{x}^{2}}-26x+5$ it is given as $\begin{align}
  & \dfrac{-\left( -26 \right)\pm \sqrt{{{\left( -26 \right)}^{2}}-4\left( 5 \right)\left( 5 \right)}}{2\left( 5 \right)}=\dfrac{26\pm \sqrt{{{26}^{2}}-100}}{10} \\
 & \Rightarrow \dfrac{26\pm \sqrt{676-100}}{10}=\dfrac{26\pm \sqrt{576}}{10} \\
 & \Rightarrow \dfrac{26\pm 24}{10}=5,\dfrac{1}{5} \\
\end{align}$
As the roots are $5,\dfrac{1}{5}$ the factors will be $\left( x-5 \right)$ and $\left( 5x-1 \right)$ .