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

2 Answers
Jan 4, 2018

Answer:

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.

Jan 4, 2018

Answer:

#(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)#