Think about 5

Simpler solution by @XerDazzle XD酱 as follows:

$(x+3)(x+4)(x+6)(x+7) = 1120 = 4 \times 5 \times 7 \times 8$ $\implies \begin{cases} x = 1 & \implies (1+3)(1+4)(1+6)(1+7) = 4 \times 5 \times 7 \times 8 \\ x = -11 & \implies (-11+3)(-11+4)(-11+6)(-11+7) = (-8)(-7)(-5)(-4) \end{cases}$

Therefore, sum of solutions is $1-11 = \boxed{-10}$ .

Previous solution

Let $y = x+5$ (see note), then:

$\begin{aligned} (x+3)(x+4)(x+6)(x+7) & = 1120 \\ \color{#D61F06}{(y-2)} \color{#3D99F6}{(y-1)}\color{#3D99F6}{(y+1)}\color{#D61F06}{(y+2)} & = 1120 \\ \color{#3D99F6}{(y^2-1)}\color{#D61F06}{(y^2-4)} & =1120 \\ y^4 - 5y^2 + 4 &= 1120 \\ y^4 - 5y^2 -1116 &= 0 \\ (y^2+31)(y^2-36) & = 0 \\ \Rightarrow y & = \pm 6 \\ \Rightarrow x & = \begin{cases} -6-5 & = -11 \\ 6-5 & = 1 \end{cases} \end{aligned}$

Therefore the sum of $x$ satisfying the equation $= -11+1 = \boxed{-10}$

Note that 5 is the arithmetic mean of 3, 4, 6 and 7. Using $y=x+5$ results in the four factors $y \pm 1$ and $y \pm 2$ , so that we eventually get a quadratic equation $y^4 - 5y^2 + 4 = 1120$ without the $y^3$ and $y$ terms, which can be easily solved.

$(x+3)(x+4)(x+6)(x+7)=1120$

$(x+3)(x+7)(x+4)(x+6)=1120$

$(x^2+10x+21)(x^2+10x+24)=1120$

Let $x^2+10x=a$

$(a+21)(a+24)=1120$

$a^2+45a+504=1120$

$a^2+45a-616=0$

$(a+56)(a-11)=0$

$\Rightarrow a = -56 \quad or \quad a = 11$

When $a=-56 \Rightarrow x^2+10x+56=0 \Rightarrow x=\dfrac{-10\pm\sqrt{-124}}{2}$ which are complex roots , hence not considered.

When $a=11 \Rightarrow x^2+10x-11=0$ , the discriminant is positive.

Hence by Vieta's formula , sum of real roots $\huge=\boxed{-10}$

We have $(x+3)(x+4)(x+6)(x+7)=1120.$ Let $x+5=z$ . The expression becomes $(z-2)(z-1)(z+1)(z+2)=1120.$

$(z^2-4)(z^2-1)=1120$

$z^4-5z^2+4=1120$

$z^4-5z^2-1116=0$

$(z^2-36)(z^2+31)=0$

The only real values of $z$ for which this equation is true is $z=6$ and $z=-6$ . $x=z-5$ so the real $x$ satisfying this equation are $x=1$ and $x=-11$ .

$1+-11=\boxed{-10}$

We factor 1120 into 7x5x2^5=1120.

With a little playing around one can easily find that one positive solution would be 4,5,7 and 8. Where x=1.

We are not done yet. It also appears there could be a negative solution also for this. We then further rule out 2 negatives and 2 positives because that cannot be consistent with the equation. Thus the only solution must have the characteristics of 4 negatives.

Then again there are two possibilities. We already know that the numbers are 4,5,7 and 8 but we don't know which 'way' the equation starts at. In other words does -4=x+3 or -8=x+3? We test the later and find that it works thus giving us x=-11.

We add 1+(-11) together and get -10, our answer.

$1120=4*5*6*7 ~~so ~if~ x+3=4~we~get~ the~ solution~x=1.. \\~Now~the~constant~term~is~by~Vieta's Formula~\\=3*4*6*7-1120=-616=- product=-7*8*11.\\~~\\If~x=\pm~7,~~(7+6)=13 ~or~(-7+4)=-3~not ~ factors~ of~~1120.\\If~x=\pm~8, ~(x+7)=15 ~not ~a~ factor~ ~1120~but~-8~is.\\If~x=\pm~11,~(11+4)=15~is~not~a~factor~of~1120~but -11~is.\\~~\\But~ with ~x=-8, ~the~ product ~of ~four~brackets~is ~not~1120.\\But~ with ~x=-11, ~the~ product ~of ~four~brackets~is~1120.\\\text{Now the sum of all roots = -(3+4+6+7)= - 20. So the sum of the }\\\text{remaining two roots is = -20 +1 - 11= -10. So other roots can be 1, and -11. }\\\text{That means the expression is a square., but the last is - tive and not a square.}\\\text{ So other two roots are not real. Sum of real roots =1 + (-11) = - 10.}$

The relation with the picture is that there is no $x+5$ term, and hence, one substitutes $x = t-5$

We have, after the substitution,

$\displaystyle (t-2)(t-1)(t+1)(t+2) = 1120 \implies (t^2 -1)(t^2 -4) = 1120$

Further substitute $s=t^2$ . Then,

$\displaystyle (s-1)(s-4) = 1120 \implies s^2 - 5s -1116 = 0$

Since we have real numbers in play here, use the Quadratic Formula, to get:

$\displaystyle s = \frac{5 \pm \sqrt{4489}}{2} = \frac{5 \pm 67}{2} \implies s = 36 \text{ or} -31$

$\displaystyle \implies t^2 = 36 \text{ or} -31 \text{ (rejected, since } t^2>0\text{)} \implies t = \pm 6$

$\displaystyle \implies x = \pm 6 - 5 = 1 \text{ or} -11$

Hence, $\displaystyle \sum_{x\in\Re} x = \boxed{-10}$

I have a non-standard but relatively quick solution to the problem.

Prime factorise 1120 and you get $2^5 × 5 ×7$ which can be rewritten as 4×5×7×8.

Note that this is exactly in the format of the LHS expression, clearly $x=1$ is a solution.

But consider $-8×-7×-5×-4$ as a possibility thus $x=-11$ is a solution too. Adding both gives you the required answer.

I found this fairly easy to be solved just by inspection.

The prime factors of 1120 are $2^{5}$ , 5 and 7, so it's easy to see that one of the solutions is x = 1. For the other solution, each of the four factors must be the negative equivalents of all the four factors (4, 5, 7 and 8), so x = -11.

Note that x cannot take any other value because each of the factors are in the from of (x + a), and so there is only one combination of real numbers that will produce any real product.

Let (x+5)=a . (a-2)(a-1)(a+1)(a+2)=1120.

As many earlier solutions have noted, we can use the prime factorization of 1120 to quickly find two solutions.

$1120 = 2^5 * 5 * 7$

$1120 = 32 * 35$

$1120 = 4 * 5 * 7 * 8 = -8 * -7 * -5 * -4$

By inspection, we see that the first factorization gives us x = 1, while the second gives us x = -11.

To see that there are no other real solutions, consider the graph of

$f(x) = (x+3) (x+4) (x+6) (x+7)$

This is a quartic with a positive leading coefficient and four zeros at -7, -6, -4 and -3 (i.e. it looks like a W.) Thus it has to intersect the line y=1120 at least twice, once to the left of the four zeros and once to the right. Any other intersections of the two graphs would have to occur between the two middle zeros (as between -7 and -6, as well as between -4 and -3, the graph lies below the x-axis.)

By symmetry, the maximum value for f(x) between the middle zeros must occur at x = -5; but f(-5) = 4, so the graphs do not intersect anywhere other than the two points mentioned earlier, and thus x = -11 and x = 1 are indeed the only real solutions to the given equation.

First, notice that $x\not\geq6$ because:

$(6+3)(6+4)(6+6)(6+7)=6^4+O(6)+3\times4\times6\times7=1296+O(6)+504>1120$

Then, after plugging in some values we find that $x=1$ works, and since there is an even number of factors, we can obtain the same result with negative numbers due to the fact that $7+3=10$ , and $6+4=10$ , which can be "inverted" to get the same numbers but with opposite sign when subtracting $10$ . And to get the same factors we subtract $1$ again, hence $x=-10-1=-11$ works too.

So the sum is: $\boxed{-11+1=-10}$

If you think of the 1120 as the area of a rectangle.

You can arrange the two sides of the rectangle as (x+3)(x+7) and (x+4)(x+6).

When you multiply out the brackets, you are left with x^2+10x+21 as one side and x^2+10x+24 as the other.

The latter can then be written as x^2+10x+(21+3).

Therefore the area of the rectangle can be found to be (x^2+10x+21)^2 + 3(x^2+10x+21) = 1120.

Then by Substituting x^2+10x+21 for y. You get y^2+3y-1120=0.

When solving this equation you get 2 solutions 32 and -35.

Then when you substitute x^2+10x+21 back in and try to solve when = to 32 and -35. The solution for when = 32 is -11 and 1 and there is no real solution for when = -35.

So the answer is -11 + 1 which is -10.

Notice that $1120 = 2^5 \times 5 \times 7$

If we compare and rearrange with the $LHS$ expression, we can see:

$(x+3)(x+4)(x+6)(x+7)= 2^3 \times 2^2 \times 5 \times 7$

$\Leftrightarrow (x+3) \times (x+6) \times (x+4) \times (x+7) = 4 \times 7 \times 5 \times 8$

This implies that we need to find $x$ such that it "fits" (or so-called 'symmetric to-') the arrangement on the $RHS$ . We can see that $\boxed{x = 1}$ is an answer. However, we must check another way that we can arrange this.

We try to attain $x$ to a negative value. By doing this we have:

$[-1(x+3)][-1(x+4)][-1(x+6)][-1(x+7)]= 2^3 \times 2^2 \times 5 \times 7$

$(-x-3)(-x-4)(-x-6)(-x-7)= 4 \times 7 \times 5 \times 8$

Eyeballing for $x$ gives us $\boxed{x = -11}$

The reason why we do this is because of these two possible factorizations of $1120$ :

• $-2^3 \times -2^2 \times 5 \times 7$

• $-2^3 \times -2^2 \times -5 \times -7$

which have negative values that could accord to the products of the terms in the $LHS$ polynomial, but gives the value of a positive integer. We see that the first option on the list is not symmetric to the $LHS$ polynomial, hence not a feasible factorization. Second option is possible because it is symmetric to the $LHS$ polynomial since its just every term in the product times $-1$ .

Therefore, the sum of all solution to $x$ is $\large \boxed{ -11 + 1 = -10}$

Simply separate 1120 into primes. Note: 5, 7, 2^2, 2^3. Since there are even number of factors two solutions exists; the above values for positive and negative. Which only occurs for 1 and -11; 1 + - 11 = -10.

By analysis, we can see that it will have 2 real roots . Let they be a and b. As product has 4 terms , negative sign gets compensated. We have , (a+3)=-(b+7) , (a+4)=-(b+6) , and so on. Solving any one, we get a+b=-10.

Probably should have solved it using algebra... but that's no fun. Instead, here's a numerical method.

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from math import *

# Newton's method for finding roots
# Original function
f = lambda x: (x+3)*(x+4)*(x+6)*(x+7)-1120
# Derivative:
fp = lambda x: (x+4)*(x+6)*(x+7)+(x+3)*((x+6)*(x+7)+(x+4)*(2*x+13))

roots={}
guesses=[]

i=-617         # Highest coefficient = 616 -> loose upper bound
while i <= 617:
    guesses.append(i)
    i+=0.5

for guess in guesses:
    val = f(guess)
    deriv = fp(guess)
    if deriv == 0:
        continue
    nextguess = -val/deriv
    err = 1e-5

    while fabs(nextguess-guess)>err and fabs(nextguess-guess)<1e5:
        guess = nextguess
        nextguess = guess-f(guess)/fp(guess)

    roots[round(nextguess, 5)] = 0

for root in roots.keys():
    print(root)

$(x+3)(x+4)(x+6)(x+7)=(x+3)(x+7)(x+4)(x+6)$

$(x+3)(x+4)(x+6)(x+7)=(x+5-2)(x+5+2)(x+5-1)(x+5+1)$

$(x+3)(x+4)(x+6)(x+7)=\left((x+5)^2-4)((x+5)^2-1\right)$

$1120 = 32\cdot 35=(-32)\cdot(-35)$

So we've obviously got 4 solutions, the two solutions of $(x+5)^2 = 36$ , which are 1 and -11, and the two solution of $(x+5)^2=-31$ which aren't real.

Since this is an equation of degree four, they are all the solutions there are.

Make the equation above becomes : ( $4)(5)(7)(8)= 1120$ or ( $-8)(-7)(-5)(-4)=1120$

Therefore ( $x=1)$ and ( $x=-11)$ So , The Answer is ( $1-11= -10)$

Consider that if (x+3)=-(y+7), x and y real roots, we have the same numbers being multiply with negative signal, but as (-1)^4=1 it gives the same number.

x+3=-y-7

x+y=-7-3

x+y=-10

take (x+5=y)....then we shall get the equation simlified as, (y^4)-(y^2)-1116=0..........now by middle term factorisation we get (y^2)=36...i.e y=6,-6....x=y-5=1,-11 therefore the sum is -10

If y=x+5

Then the equation became

(y-2)(y-1)(y+1)(y+2)=1120

(y-2)(y+2) (y-1)(y+1)=(y-2^2)(y^2-1^2)

y^4-5y^2+4=1120

Bhaskara for y^2

y^2=36 or y^2=-31

Only real solutions

y=6 or y=-6

So x=1 or x=-11

Finally

-11+1=-10