Question

How do you factor${x^3} - 2{x^2} - 35x$?

Answer 1

Hint:
Here we can take the $x$ common and then we will get one quadratic equation which we can factorise by completing the square method in which 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 equation which is ${x^3} - 2{x^2} - 35x$
As $x$ is common in all the terms we can take it common and write the above cubic equation as $x\left( {{x^2} - 2x - 35} \right)$$ - - - - (1)$
Now we just need to solve the quadratic equation which is ${x^2} - 2x - 35$$ - - - - (2)$
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 ${x^2} - 2x - 35$ that the coefficient of $x{\text{ is }} - 2$
So we need to add and subtract the value ${\left( {\dfrac{{ - 2}}{2}} \right)^2} = 1$ in the above quadratic equation which is given. Hence we will get:
${x^2} - 2x - 35 + 1 - 1$
Now we can write it in the form:
$
  {x^2} - 2x + 1 - 35 - 1 \\
  ({x^2} - 2x + 1) - (35 + 1) \\
 $
Now we know that we can write the term in the first bracket of the equation (1) as:
$({x^2} - 2x + 1) = {\left( {x - 1} \right)^2}$
Now we can substitute this square calculated values in the equation (2) and we will get:
${x^2} - 2x - 35 = $${\left( {x - 1} \right)^2} - 36$
Now we can write $36 = {6^2}$
${\left( {x - 1} \right)^2} - {\left( 6 \right)^2}$
Now we know that:
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
So we can apply it over here and get:
${\left( {x - 1} \right)^2} - {\left( 6 \right)^2}$
$\left( {x - 1 + 6} \right)\left( {x - 1 - 6} \right)$
On simplifying it we will get:
$\left( {x + 5} \right)\left( {x - 7} \right)$
Hence we get the quadratic equation after factoring as $\left( {x + 5} \right)\left( {x - 7} \right)$
Hence now we have got ${x^2} - 2x - 35 = $$\left( {x + 5} \right)\left( {x - 7} \right)$
Now substituting this value in equation (1) we will get:
$x\left( {x + 5} \right)\left( {x - 7} \right)$

Hence we get the result as $x\left( {x + 5} \right)\left( {x - 7} \right)$

Note:
Here we can do it even by splitting the $ - 2x$ in the quadratic equation as $ - 7x + 5x$
Then we will get:
${x^2} - 2x - 35$
$
  {x^2} - 7x + 5x - 35 \\
  x\left( {x - 7} \right) + 5\left( {x - 7} \right) \\
  \left( {x + 5} \right)\left( {x - 7} \right) \\
 $
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.