Question

How do I factorise (x+1)(x+2)(x+3)(x+4)-120 ?

Answer 1

For this one I will merely give the answer as `(x-a)(x-b)(x^2+cx+d)`. Use what is below to work the values out.

Explanation:

Multiplying out we have:

`(x+1)(x+2)(x+3)(x+4)-120=x^4+10x^3+35x^2+50x-96`

Can you find a pair of rational (actually, integer) zeroes of this polynomial? Hint:

`120=2×3×4×5=(-5)×(-4)×(-3)×(-2)`

When you get these roots, they become the values of `a` and `b` in the above answer. Beware that one of these is negative. Divide by those two factors `x-a`, `x-b` and you are left with a quadratic quotient `x^2+cx+d` that has no more real roots (because the discriminant `c^2-4d` is negative).

Factoring is as much an art as a science.

Answer 2

`(x^2+5x+16)(x+6)(x-1)`

Explanation:

`(x+1)(x+2)(x+3)(x+4)-120`

=`(x^2+5x+4)*(x^2+5x+6)-120`

=`(x^2+5x+5)^2-1-120`

=`(x^2+5x+5)^2-11^2`

=`(x^2+5x+5+11)*(x^2+5x+5-11)`

=`(x^2+5x+16)*(x^2+5x-6)`

=`(x^2+5x+16)(x+6)(x-1)`