Question

How do you factor the expression ${x^2} - 10x + 25$?

Answer 1

Hint: Whenever we need to factorise any quadratic equation we need to add and subtract the value of ${\left( {\dfrac{{{\text{coefficient of }}x}}{2}} \right)^2}$ in the given quadratic equation. Now we will get the equation in form of the ${\left( {a + b} \right)^2} \pm {\text{c or }}{\left( {a - b} \right)^2} \pm {\text{c }}$ and now we can simplify it further as much as we can.

Complete step by step solution:
Here we are given the quadratic equation which we need to factorize. So we must know that any equation of degree $2$ is known as the quadratic equation. In order to find its factor we need to add and subtract the value of ${\left( {\dfrac{{{\text{coefficient of }}x}}{2}} \right)^2}$ in the given quadratic equation.
So here we can see in the above equation which is ${x^2} - 10x + 25$ that the coefficient of $x{\text{ is }} - 10$.
So we need to add and subtract the value ${\left( {\dfrac{{ - 10}}{2}} \right)^2} = 25$ in the above quadratic equation which is given. But we already have this term here. Here the addition term must be there necessarily. So now we do not need to add and subtract. Hence this means that this is the square of any term.
${x^2} - 10x + 25$
Now we can write it in the form:
${x^2} - 2\left( 5 \right)\left( x \right) + {\left( 5 \right)^2}$
We also know that ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Now we can compare ${x^2} - 2\left( 5 \right)\left( x \right) + {\left( 5 \right)^2}$ with ${a^2} - 2ab + {b^2}$
We can say that:
$
  a = x \\
  b = 5 \\
 $

Hence we can compare and say that ${x^2} - 2\left( 5 \right)\left( x \right) + {\left( 5 \right)^2}$$ = {\left( {x - 5} \right)^2}$

Note:
Here we can do it even by splitting the $ - 10x$ in the quadratic equation as $ - 5x - 5x$
Then we will get:
${x^2} - 10x + 25$
$
  {x^2} - 5x - 5x + 25 \\
  x\left( {x - 5} \right) - 5\left( {x - 5} \right) \\

  \left( {x - 5} \right)\left( {x - 5} \right) = {\left( {x - 5} \right)^2} \\
 $
Hence this method can also be used over here but the method we used is applicable always when we do not know how to split the coefficient of $x$ term.