SOLUTION: Factorise completely 16x^4 � 40x^2 + 9

Question 167164: Factorise completely 16x^4 � 40x^2 + 9
Answer by J2R2R(94) (Show Source): You can put this solution on YOUR website!
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)
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