Question

(x-3)(x-4)(x-5)(x-6)=1680

Answer 1

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Answer 2

To solve the equation $(x-3)(x-4)(x-5)(x-6) = 1680$, we can follow these steps:

Step 1: Expand the equation Step 2: Set it to zero and rearrange Step 3: Solve for $x$

Let's proceed with these steps:

Step 1: Expand the equation Expand the left side of the equation $(x-3)(x-4)(x-5)(x-6)$ using the FOIL method (First, Outer, Inner, Last):

$(x-3)(x-4)(x-5)(x-6) = (x^2 - 7x + 12)(x^2 - 11x + 30)$

Step 2: Set it to zero and rearrange Now we have:

$(x^2 - 7x + 12)(x^2 - 11x + 30) = 1680$

Subtract 1680 from both sides to set the equation to zero:

$(x^2 - 7x + 12)(x^2 - 11x + 30) - 1680 = 0$

Step 3: Solve for $x$ Now, we need to find the solutions for $x$. We can use various methods, but in this case, we can try to factorize the equation further:

$(x^2 - 7x + 12)(x^2 - 11x + 30) - 1680 = (x^2 - 7x + 12 - 40)(x^2 - 11x + 30) = (x^2 - 7x - 28)(x^2 - 11x + 30) = 0$

Now, we have two quadratic equations:

$x^2 - 7x - 28 = 0$ and $x^2 - 11x + 30 = 0$

Let's solve them one by one:

For $x^2 - 7x - 28 = 0$:

To factorize, we look for two numbers that multiply to give -28 and add up to -7. Those numbers are -4 and 7.

So, $x^2 - 7x - 28 = (x - 4)(x + 7) = 0$

Setting each factor to zero:

$x - 4 = 0 \Rightarrow x = 4$

$x + 7 = 0 \Rightarrow x = -7$

For $x^2 - 11x + 30 = 0$:

To factorize, we look for two numbers that multiply to give 30 and add up to -11. Those numbers are -5 and -6.

So, $x^2 - 11x + 30 = (x - 5)(x - 6) = 0$

Setting each factor to zero:

$x - 5 = 0 \Rightarrow x = 5$

$x - 6 = 0 \Rightarrow x = 6$

Now we have four potential solutions for $x$: $x = 4$, $x = -7$, $x = 5$, and $x = 6$.

However, we should check whether any of these solutions make the original equation undefined (e.g., division by zero) or result in the left-hand side being equal to zero.

Let's plug each value back into the original equation $(x-3)(x-4)(x-5)(x-6) = 1680$ and see which ones satisfy the equation:

1. $x = 4$: $(4-3)(4-4)(4-5)(4-6) = 1 * 0 * (-1) * (-2) = 0 \neq 1680$

2. $x = -7$: $(-7-3)(-7-4)(-7-5)(-7-6) = (-10) * (-11) * (-12) * (-13) = 17160 \neq 1680$

3. $x = 5$: