Question

Find the factors of \[18{x^2} - 2\].

Answer 1

Hint:A difference of two perfect squares can be factored into. Commutative property of multiplication can be used here. Pull out the like factors that are the like terms for getting the factors easily. Each term must be solved equal to zero separately. As many terms are there in the product, those many equations must be solved.

Complete step by step solution:
Pulling out the like factors we have,
Factoring as a difference of squares we have,
\[ \Rightarrow \left( {9{x^2} - 1} \right)\]
A difference of two perfect squares \[\left( {{a^2} - {b^2}} \right)\]can be factored into.
Proof:
\[
\left( {a + b} \right) \times \left( {a - b} \right) \\
\Rightarrow {a^2} - ab + ba - {b^2} \\
\Rightarrow {a^2} - ab + ab - {b^2} \\
\Rightarrow {a^2} - {b^2} \\
\]
Where, \[ab = ba\] is the commutative property of multiplication and \[ - ab + ab\] equals to zero and is therefore eliminated from the expression.
Since, 9 is the square of 3, 1 is the square of 1 and \[{x^2}\] is the square of \[{x^1}\]therefore
The factorization is: \[\left( {3x + 1} \right)\left( {3x - 1} \right)\]
Now the equation can be written as:
Since, a product of several terms equals zero.
When a product of two or more terms equals zero, then a minimum of one among the terms must be zero.
We shall now solve each term equals zero separately. In other words we are getting to solve as many equations as there are terms within the product.
Solving \[2 = 0\]has no solution. A non-zero constant never equals to zero.
Solving \[\left( {3x + 1} \right) = 0\],
Subtracting 1 from both sides of the equation:
\[3x = - 1\]
Dividing both the sides of the equation by 3:
\[x = - \dfrac{1}{3} = - 0.33\]
Solving\[\left( {3x - 1} \right) = 0\],
Adding 1 to both sides of the equation,
\[3x = 1\]
Dividing both the sides of the equation by 3:
\[x = \dfrac{1}{3} = 0.33\]
Hence two solutions are \[x = \dfrac{1}{3}\]and\[x = - \dfrac{1}{3}\].

Note: Any solution of term \[ = 0\] solves product \[ = 0\] as well. The number of terms which are present that is equal to the number of equations to be solved. If the number of terms and equations doesn’t match then there is some mistake in calculations. Number of equations can’t be more than the number of terms and vice versa.