Question

Factorise the following: $ 3{a^2} - 11ab + 6{b^2} $
A. $ \left( {a - 2b} \right)\left( {a - b} \right) $
B. $ \left( {3a + b} \right)\left( {5a - b} \right) $
C. $ \left( {a - 2b} \right)\left( {2a - 3b} \right) $
D. $ \left( {3a - 2b} \right)\left( {a - 3b} \right) $

Answer 1

Hint: In this expression we can find the factors by using the method first check the end terms if they are perfect squares then solve the expression by using the identity of $ {\left( {a - b} \right)^2} $ . When the end terms of the expression is not a perfect square then solves the expression by using the splitting method.

Complete step-by-step answer:
The given expression is $ 3{a^2} - 11ab + 6{b^2} $ .
Dividing the whole expression by 3 then we get,
 $
 \Rightarrow \dfrac{{3{a^2}}}{3} - \dfrac{{11ab}}{3} + \dfrac{{6{b^2}}}{3}\\
 \Rightarrow {a^2} - \dfrac{{11ab}}{3} + 2{b^2}\;
 $
The given expression has the degree of 2, as the highest exponent in the expression is 2. Now we check the end terms of the expression which are $ {a^2} $ and $ 2{b^2} $ .
Here, we found that $ {a^2} $ is the perfect square of a but $ 2{b^2} $ is not a perfect square so; we will use a splitting method to solve the expression.
In the splitting method we have to split the middle term into two parts in this way that make the values common and factors can carry out easily.
So, in the expression $ {a^2} - \dfrac{{11ab}}{3} + 2{b^2} $ , we will split the middle term $ \dfrac{{11}}{3} $ into two numbers which on addition give $ \dfrac{{11}}{3} $ back and on multiplying give 2,
Hence 3 and $ \dfrac{2}{3} $ are two terms which satisfy the given condition so substituting the values in place of 10 then we get,
$ \Rightarrow {a^2} - \left( {3 + \dfrac{2}{3}} \right)ab + 2{b^2}$
Now, multiplying p with 3 and $ \dfrac{2}{3} $ to open the bracket,
$\Rightarrow {a^2} - 3ab - \dfrac{2}{3}ab + 2{b^2}$
We found that in first two terms a is common so taking a common from first two terms and -2b from last two terms then,
 $
 \Rightarrow a\left( {a - 3b} \right) - 2b\left( {\dfrac{1}{3}a - b} \right)\
 \Rightarrow a\left( {a - 3b} \right) - 2b\left( {\dfrac{{a - 3b}}{3}} \right)\
 \Rightarrow a\left( {a - 3b} \right) - \dfrac{{2b}}{3}\left( {a - 3b} \right)\;
 $
So we get same values under the brackets taking the bracket values as common and the remaining part in other bracket then,
$
 \Rightarrow \left( {a - 3b} \right)\left( {a - \dfrac{{2b}}{3}} \right)\\
 \Rightarrow \left( {a - 3b} \right)\left( {\dfrac{{3a - 2b}}{3}} \right)\\
 \Rightarrow \left( {a - 3b} \right)\left( {3a - 2b} \right)\;
 $
Both expressions are linear hence they are the simplest form and the factorise form of the given expression.
So, the correct answer is “Option D”.

Note: A quadratic polynomial is a polynomial of degree 2. A univariate quadratic polynomial has the form. . An equation involving a quadratic polynomial is called a quadratic equation. A closed-form solution known as the quadratic formula exists for the solutions of an arbitrary quadratic equation