# Math Help - hard equation !

1. ## hard equation !

HI !!

How can I solve this equation ???

$(x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \$

2. 1. Expand

2. Set equal to 0

3. Factorise if possible and solve using NFL.

3. Here's a clever little shortcut. 120 is equal to 5!. So x = 1 works.

4. Originally Posted by yehoram
HI !!

How can I solve this equation ???

$(x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \$
Dear yehoram,

$(x+1)(x+2)(x+3)(x+4) = 120=2\times 3\times 4\times 5$

Therefore it is clear that x=1 is a solution.

$(x+1)(x+2)(x+3)(x+4)-120=(x-1)(Ax^3+Bx^2+Cx+D)=0$

Evaluating A,B,C and D will give you a cubic equation. The solutions can be obtained using several methods such as the Cardano's method. Refer: Cubic function - Wikipedia, the free encyclopedia

5. The key to this problem is to partition the products into two : (1) the first and the last , (2) the 2nd and the 3rd , ie

$(x+1)(x+2)(x+3)(x+4) = 120$

$[(x+1)(x+4)][(x+2)(x+3)] = 120$

$(x^2 + 5x + 4 )(x^2 + 5x + 6) = 120$

Then sub. $u= x^2 + 5x + 5$ so we have

$(u-1)(u+1) = 120$ or

$u^2 = 121$ thus

$u = 11$ or $u = -11$

$x^2 + 5x - 6 = 0$ or $x^2 + 5x + 16 = 0$

The four solutions are

$-6,1 , \frac{-5+\sqrt{39} i }{2} , \frac{-5-\sqrt{39} i }{2}$

6. Hello, yehoram!

How can I solve this equation?

. . $(x+1)(x+2)(x+3)(x+4) \:=\: 120$

We have: the product of four "consecutive" numbers equals 120.

The only candidates are: . $\{2,3,4,5\}\,\text{ and }\,\{\text{-}5,\text{-}4,\text{-}3,\text{-}2\}$

7. Originally Posted by yehoram
HI !!

How can I solve this equation ???

$(x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \$
Along the same lines as Dr Steve's "zwischenzug", what if $x=-6\;\;?$

(my answer looks a bit silly now. What's with the clock? Soroban's response wasn't there when I started!)

8. Originally Posted by yehoram
$(x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \$
x^4 + something + 24 = 120
x^4 = 96 - something
x^4 < 96

Well?!