Question

Factorise by middle splitting${x^2} - 22x + 120$ .

Answer 1

Hint: t: The given expression can be factorise by using the identity of $ {\left( {a - b} \right)^2} $ if the end terms of the expression are the perfect square otherwise they can be factorise by splitting the middle term of the expression.

Complete step-by-step answer:
The given expression for the factorization is: $ {x^2} - 22x + 120 $
Here we see the end terms of the given expression which are $ {x^2} $ and 120. So we found that $ {x^2} $ is the perfect square of x but 120 is not the perfect square.
Therefore, we have to use the splitting method to solve the expression.
We have to replace the middle term 22 or we have to distribute the middle term 22 in two parts so that their addition will be equal to 22 and their multiplication is equal to the 120.
So 10 and 12 are the number or digit which can be the middle term 10 then we get,
 $ \begin{array}
 \Rightarrow {x^2} - \left( {10 + 12} \right)x + 21\\
 \Rightarrow {x^2} - 10x - 12x + 120
\end{array} $
Here we see that x is common between the first two terms of the expression and 12 is common between the last two terms of the expression. So carry out these common then we get,
 $ \Rightarrow x\left( {x - 10} \right) - 12\left( {x - 10} \right) $
Take common $ \left( {x - 10} \right) $ from the expression and the remaining expression in the other bracket.
 $ \Rightarrow \left( {x - 10} \right)\left( {x - 12} \right) $
So, this expression is the required factor of the given expression. $ \left( x-10 \right)\left( x-12 \right) $ are the factors of the given expression.

Note: The factor of the quadratic expression cube carried out by using the identities and also by using the splitting middle term method. Identities can be applicable for equations of higher degree but the splitting method is only applicable when the expression has quadratic form.