Question

Factorise:
(i) $ 4{x^2} + 9{y^2} + 16{z^2} + 12xy - 24yz - 16xz $
(ii) $ 2{x^2} + {y^z} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $

Answer 1

Hint: Best way of solving the above questions will be by using the expansion formula of $ {(a + b + c)^2} $ . Write, the above mentioned equations in the form of the expansion formula of $ {(a + b + c)^2} $ to factorize them.

Complete step-by-step answer:
(i) $ 4{x^2} + 9{y^2} + 16{z^2} + 12xy - 24yz - 16xz $
We have the formula,
 $ {(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca $ . . . (1)
We shall try to convert the expression in the question in the form of RHS of equation (1).
By arranging the expression in the question, we can write it as
\[{(2x)^2} + {(3y)^2} + {(4z)^2} + 12xy - 24yz - 16xz\]
\[ = {(2x)^2} + {(3y)^2} + {( - 4z)^2} + 2(2x)(3y) + 2(3y)( - 4z) + 2(2x)( - 4z)\] . . . (2)
Now, comparing equation (2) with the RHS of equation (1), we get
 $ a = 2x,b = 3y,c = - 4z $
Therefore, by substituting these values in the LHS of equation (1), we can simplify the expression of equation (2) as,
 $ {(2x)^2} + {(3y)^2} + {( - 4z)^2} + 2(2x)(3y) + 2(3y)( - 4z) + 2(2x)( - 4z) = {(2x + 3y - 4z)^2} $
 $ = (2x + 3y - 4z)(2x + 3y - 4z) $
Hence, $ 4{x^2} + 9{y^2} + 16{z^2} + 12xy - 24yz - 16xz $ can be factorized as $ (2x + 3y - 4z)(2x + 3y - 4z) $
(ii) $ 2{x^2} + {y^z} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $
We know that,
 $ {(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac $ . . . . . (3)
We shall try to convert the expression in the question in the form of RHS of equation (1).
By arranging the expression in the question, we can write it as
 $ {\left( {\sqrt 2 x} \right)^2} + {y^z} + {(2\sqrt 2 z)^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $
 $ = {\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2( - \sqrt 2 x)y + 2y(2\sqrt 2 z) + 2(2\sqrt 2 z)( - \sqrt 2 x) $ . . . (4)
Now, comparing equation (4) with the RHS of equation (3), we get
 $ a = - \sqrt 2 x,b = y,c = 2\sqrt 2 z $
Therefore, by substituting these values in the LHS of equation (1), we can simplify the expression of equation (2) as,
 $ {\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2( - \sqrt 2 x)y + 2y(2\sqrt 2 z) + 2(2\sqrt 2 z)( - \sqrt 2 x) = {( - \sqrt 2 x + y + 2\sqrt 2 )^2} $
 $ = \left[ {( - \sqrt 2 x) + y + (2\sqrt 2 z)} \right]\left[ {( - \sqrt 2 x) + y + (2\sqrt 2 z)} \right] $
Hence, $ 2{x^2} + {y^z} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $ can be factorized as $ \left[ {( - \sqrt 2 x) + y + (2\sqrt 2 z)} \right]\left[ {( - \sqrt 2 x) + y + (2\sqrt 2 z)} \right] $

Note: You need to write the given expressions in the standard form of the formula to be able to compare the corresponding terms. For that, you need to make sure that the negative terms are arranged accordingly, just like we did in this question. Do not randomly add negative signs to any term. The negative sign must justify its position.