Question

Factorise:
\[{{p}^{2}}-3pq+2{{q}^{2}}\]
(a) \[\left( 3p-2q \right)\left( p+q \right)\]
(b) \[\left( 3p-2q \right)\left( p-q \right)\]
(c) \[\left( p-2q \right)\left( p-q \right)\]
(d) \[\left( p-q \right)\left( p-q \right)\]

Answer 1

Hint: In this question, we first need to know about the solution of a quadratic equation. Then to find the solution of a quadratic equation we use different methods in which factorisation method is one of them. We can factorise the above equation by using the formula from factorisation method.

Complete step-by-step answer:
\[a{{x}^{2}}+bx+c=a\left( x-\alpha \right)\left( x-\beta \right)\]
Let us look at some of the basic definitions first.
Let us look into some of the definitions and results of polynomials.
POLYNOMIAL- An expression of the form \[{{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+......+{{a}_{n-1}}x+{{a}_{n}},\] where \[{{a}_{0}},{{a}_{1}},{{a}_{2}}.......,{{a}_{n-1}},{{a}_{n}}\] are real numbers and n is a non-negative integer, is called a polynomial in the variable x. Polynomials in variable x are generally denoted by \[f\left( x \right)\].
QUADRATIC POLYNOMIAL- A polynomial having degree two.
QUADRATIC EQUATION- A quadratic polynomial f(x) when equated to zero is called quadratic equation.
\[\Rightarrow a{{x}^{2}}+bx+c=0\], where \[a\ne 0\]
Roots of a Quadratic Equation: The values of the variable x which satisfy the quadratic equation are called roots of quadratic equation.
Solution of a Quadratic Equation by Factorisation Method-
Let \[a{{x}^{2}}+bx+c=a\left( x-\alpha \right)\left( x-\beta \right)=0\]. Then, \[x=\alpha \] and \[x=\beta \] will satisfy the given equation.
Where \[a\left( x-\alpha \right)\left( x-\beta \right)\] is the factorisation of \[a{{x}^{2}}+bx+c\].
Now, the given equation in the question can be considered as a quadratic equation in p.
Thus, it can be factored as:
\[\begin{align}
  & \Rightarrow {{p}^{2}}-3pq+2{{q}^{2}} \\
 & \Rightarrow {{p}^{2}}-pq-2pq+2{{q}^{2}} \\
\end{align}\]
Now, let us take the common terms out and rewrite the equation.
\[\Rightarrow p\left( p-q \right)-2q\left( p-q \right)\]
\[\Rightarrow \left( p-q \right)\left( p-2q \right)\]
Hence, the correct option is (c).

So, the correct answer is “Option (c)”.

Note: Instead of writing 3pq as a sum of 2pq and 1pq we can also write it in different ways. But, if written in any other form we cannot get the common terms with the other to square terms which in return does not give any further simplification. So, the factorisation may not be possible in that way.
For example if we write the term 3pq as subtraction of 4pq and 1pq then we cannot get the common terms on the both sides of arithmetic operations. So, we cannot express it as the product of two terms.
\[\begin{align}
  & \Rightarrow {{p}^{2}}-3pq+2{{q}^{2}} \\
 & \Rightarrow {{p}^{2}}+pq-4pq+2{{q}^{2}} \\
 & \Rightarrow p\left( p+q \right)-2q\left( 2p-q \right) \\
\end{align}\]
This cannot be further simplified to be written as a product of two terms as given in the options above. Thus, we need to check for the only possible way while doing the factorisation.