Question

Show that (x-1) is the factor of $ ({x^3} - 7{x^2} + 14x - 8)$ . Hence, completely factorise the given expression.

Answer 1

Hint: In this problem cubic expression is given. It is asked to find out if the given (x-1) is a factor of the equation or not. Basically, when (x-1) factorises the equation completely then the remainder will be zero. So using long division we can find other roots.
Factorization: Factors of a number are numbers that divide evenly into another number. Factorization writes a number as the product of smaller numbers.

Complete step by step solution:
Given, expression $ ({x^3} - 7{x^2} + 14x - 8)$ .
TO check whether (x-1) is a factor of the given equation or not we need to equate the factor to zero.
It means (x-1) =0
Here it gives x=1.
Putting x=1 in the given expression we get
$
   = ({x^3} - 7{x^2} + 14x - 8) \\
   \Rightarrow {(1)^3} - 7{(1)^2} + 14(1) - 8 \\
   \Rightarrow 1 - 7 + 14 - 8 = 0 \\
 $
This shows that (x-1) is the factor of $ ({x^3} - 7{x^2} + 14x - 8)$ .
Now by using long division method

Now we have two factors $ (x - 1)\,and\,({x^2} - 6x + 8)$
To factorise completely, Let
$
  f(x) = ({x^3} - 7{x^2} + 14x - 8) \\
   \Rightarrow (x - 1)({x^2} - 6x + 8) \\
   \Rightarrow (x - 1)({x^2} - 2x - 4x + 8) \\
 $
By further simplifying we get
$
  f(x) = (x - 1)[x(x - 2) - 4(x - 2)] \\
   \Rightarrow (x - 1)(x - 2)(x - 4) \\
 $
It is the complete factorisation of the given problem.

Note:
> Try to read the problem stepwise. Follow the procedure of long division step by step to get the solution. This is the prominent and errorless way to solve the problem.
> Retrace the solution to check if it is correct or not. We can equate all the terms to zero and put it in the given expression and get the idea of roots x = 1, 2, 4.
> We can also solve this question by using trial and error methods. we can assume some value for x at which the whole equation will be zero. then we can find all values of x and then form an equation by using them.