Question

Factorize the expression: \[9{a^2} + 3a - 8b - 64{b^2}\]
\[\left( 1 \right)\] \[\left( {3a - 8b} \right)\left( {3a + 8b + 1} \right)\]
\[\left( 2 \right)\] \[\left( {3a - 8b} \right)\left( {a + 8b + 1} \right)\]
\[\left( 3 \right)\] \[\left( {3a - 8b} \right)\left( {3a + b + 1} \right)\]
\[\left( 4 \right)\] \[\left( {3a - 8b} \right)\left( {3a + 8b + 9} \right)\]

Answer 1

Hint: We have to factorize the given expression \[9{a^2} + 3a - 8b - 64{b^2}\]. We solve this question using the concept of factorisation of a polynomial equation. We solve this question using the various formulas of factorisation of terms. First we will represent the given polynomial equation such that we can get terms common from it and we can apply the formula of the polynomial equations. Applying the formula and taking the terms common we will get the required factorisation of the given expression.

Complete step-by-step solution:
Given :
\[9{a^2} + 3a - 8b - 64{b^2}\]
We can also represent the equation as :
\[9{a^2} + 3a - 8b - 64{b^2} = {(3a)^2} - {(8b)^2} + 3a - 8b\]
Also, we know that \[{a^2} - {b^2} = (a + b)(a - b)\]
Using this formula we get the expression as :
\[9{a^2} + 3a - 8b - 64{b^2} = (3a - 8b)(3a + 8b) + 3a - 8b\]
Now, taking \[\left( {3a - 8b} \right)\] common from the above expression we get the expression as :
\[9{a^2} + 3a - 8b - 64{b^2} = (3a - 8b)(3a + 8b + 1)\]
Hence, the factorisation of the given expression \[9{a^2} + 3a - 8b - 64{b^2}\] is \[\left( {3a - 8b} \right)\left( {3a + 8b + 1} \right)\].
Thus, the correct option is \[\left( 1 \right)\].
If we have a complex equation such as that with power \[4\] polynomial equation, then for the factorisation of the equation using the formula as that used above may not be feasible enough so, we use a hit and trial method to find the factors of the equation. We will put the value in the given equation and it satisfies the equation then it is a factor. Then we will divide the equation with the factor and again use the same process.

Note: In each and every question which states to factorize a given expression we simply have to find the factors of the given expression. In the above question as we had two variables so on factorisation of the terms we obtained two equations as factors of the given expression. If it would have been an equation with only one variable with any order of polynomial equation then we would have obtained the solutions or the factors of the given expression.