Question

$(x - 1)(x - 2)(x + 3)(x + 4) + 6$
(Solve in Factorisation)​

Answer 1

Step-by-step explanation:

(x-1)(x-2)(x+3)(x+4)+6

(x^2+2x-1x+2)(x^2+4x+3x12)+6

(x^2+x+2)(x^2+7x+12)+6

Hope it helps you

Answer 2

Step-by-step explanation:

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

Algebra

2 Answers



Oscar L.

Jan 4, 2018

For this one I will merely give the answer as (x−a)(x−b)(x2+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=x4+10x3+35x2+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 x2+cx+d that has no more real roots (because the discriminant c2−4d is negative).

Factoring is as much an art as a science........... .

kya tum mere sath chating par bat karogi / karoge please give the answer....... .