Example Question 4

Example corresponding to question 4 in algebra_4.

Example 1


The complete factorization of $24x^4-6x^2$ is

  1. $x^2(24x^2-6)$


  2. $6x^2(4x^2-1)$


  3. $6x^2(2x-1)(2x+1)$


  4. $6(4x^4-x^2)$


  5. $x(24x^3-6x)$

Solution


  1. Complete factorization means no more factoring is possible. Start by factoring coefficients

  2. $24x^4-6x^2=6(4x^4-x^2)$

  3. Factor common terms with the variable to the same power

    $=6x^2(4x^2-1)$

  4. We now look for any factoring inside the brackets. The term $4x^2-1$ could be written as ((2x)^2 - 1^2) which is the difference of two squares and can therefore be factored.(Note: $a^2-b^2=(a-b)(a+b)$). Hence

    $=6x^2((2x)^2-1^2)= 6x^2(2x-1)(2x+1)$

  5. Note that there is no more factoring that can be done and therefore we have completely factored the given expression.
    Answer C