Factorising f(x) Completely: Find a & b Values

[denian]

given that f(x) = x^4 - 27x^2 - 14x + 120 can be expressed as

( x^2 + a )^2 - ( bx + 7 )^2

where a,b are constant. find the values of a and b. hence, or otherwise, factorise f(x) completely.




the value of a and b are -13 and 1 respectively.
so,

f(x) = ( x^2 - 13 )^2 - ( x + 7 )^2

one way to factorise f(x) completely is to substitute value of x so that we can get
f(x) = 0.

is there any other easier way since there is a keyword "HENCE" in the question.
thank you.

 

[arcnets]

f(x) = 0
<=>
( x^2 - 13 )^2 = ( x + 7 )^2
Taking square roots on both sides:
|x^2 - 13| = |x + 7|
This breaks up into 2 separate equations which are just quadratic and, thus, can directly be solved, giving 2 zeroes of f each.

 

[HallsofIvy]

I suspect what was intended here was that you use

a²- b²= (a-b)(a+b).

Once you know that
x⁴ - 27x² - 14x + 120= ( x² - 13 )² - ( x + 7 )²,

you can continue as
(x²- 13+ (x- 7))(x²-13-(x-7))
= (x²+ x- 20)(x²-x- 6)

Now can you further factor those two factors?

 

[HallsofIvy]

"That's Quaint"? :)