Question

Factorise by splitting the middle term: $ 9{(x - 2y)^2} - 4(x - 2y) - 13 $ .

Answer 1

Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factorising an expression is to take out any common factors which the terms have. So if we were asked to factor the expression $ {x^2} + x $ , since $ x $ goes into both terms, we would write $ x(x + 1) $ . We know that for the factorisation of quadratic polynomials of the form $ {x^2} + bx + c $ we have to find numbers $ p $ and $ q $ such that $ p + q = b $ and $ pq = c $ . This is called factorisation by splitting the middle term.

Complete step by step solution:
Here we have an equation: $ 9{(x - 2y)^2} - 4(x - 2y) - 13 $ . We have to factorise it by splitting the middle term.
So let us assume $ (x - 2y) = a $ . By putting this in place of $ (x - 2y) $ , we can write the equation as $ 9{a^2} - 4a - 13 $ .
Now we will solve by splitting the middle term. Here we have to split the middle term i.e. $ ( - 4a) $ in such numbers that the product of the numbers will be equal to $ - 117{a^2} $ . We can write $ - 4a = 9a - 13a $ as $ ( - 13a)(9a) = - 117{a^2} $ .
Therefore $ 9{a^2} - 4a - 13 = 9{a^2} + 9a - 13a - 13 $ , Now take out the common factor and simplify it;
 $ 9a(a + 1) - 13(a + 1) \Rightarrow (9a - 13)(a + 1) $ .
Now we put the value $ = x - 2y $ back in the equation, $ [9(x - 2y) - 13][(x - 2y) + 1] $ .
By breaking the bracket we can write it as $ (9x - 18y - 13)(x - 2y + 1) $ .
Hence the required answer is $ (9x - 18y - 13)(x - 2y + 1) $ .
So, the correct answer is “ $ (9x - 18y - 13)(x - 2y + 1) $ ”.

Note: We should keep in mind while solving this kind of middle term factorisation that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give the wrong answer. Also we should always check for the sum and product and also verify the factors by multiplying that as it will provide the same above quadratic equation or not. These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeros.