Fill in the blanks to make the equation true. ( _x + _) ( _x + _) = _x 2 + _x + _(− 35, − 8, − 5, 1, 3, 4, 6, 7, 12)

Lilliam0216 Jun 9, 2024

#1**+1 **

For any a, b, c, d:

\((ax+b)(cx+d) = acx^2 + bcx + adx + bd\)

Now a, c, and ac are all a blank in the equation. This means that a and c may only be certain numbers if the product ac is in the number bank also. The same logic also works with b and d. The only sets of numbers that have this property are (-5, 7) (product is -35) and (3, 4) (product is 12). One of these is (a,c), and the other is (b,d), but we don't know which is which. We can also flip a and b around, to get two more possibilities per equation. This means that there are 4 possibilities, and we can try out each one.

\((-5x + 3)(7x + 4) = -35x^2+(1)x+12\)

\((-5x + 4)(7x + 3) = -35x^2 + 14x + 12\)

\((3x-5)(4x+7) = 12x^2+(1)x-35\)

\((4x-5)(3x+7) = 12x^2 + 14x - 35\)

Looking at these equations, the first and third are the only ones that use only numbers in the bank. Therefore, the two solutions are **"(-5x+3)(7x+4) = -35x^2 + 1x + 12" and "(3x-5)(4x+7) = 12x^2 + 1x - 35"**

Please let me know if there is an easier way to do this!

Maxematics Jun 9, 2024