Question 167164
Factorise 16x^4 – 40x^2 + 9


There is a quick way to solve this using the (–b+-sqrt(b^2-4ac))/2a but to solve it by trial and error is better.


We have the coefficients of x^2 which when multiplied together gives 16 and the constants when multiplied together to gives 9. But the sum of the cross products i.e. the first coefficient of x^2 with the second constant plus the second coefficient of x^2 with the first constant equals -40.


What first comes to mind is 4 times 9 and 4 times 1, so for the sum to be -40 we have minus 9 and minus 1.


So (4x^2 – 9)(4x^2 – 1) = 16x^4 – 36x^2 – 4x^2 + 9 = 16x^4 – 40x^2 + 9


This is a quadratic equation in x^2 rather than x.


Now 4x^2 – 9 is the difference of two squares (2x + 3)(2x – 3)


And 4x^2 – 1 is the difference of two squares (2x + 1)(2x – 1)


Therefore


16x^4 – 40x^2 + 9 = (2x + 3)(2x – 3)(2x + 1)(2x – 1)