Question

Demanding Dwight now wants FIVE solutions to the following equation. Can you provide him with five solutions? (x – 10)(2x + 6)(x2 – 36)(x2 + 10) ( 10x + 20) = 0 Do you think there might be a sixth solution?

Answer 1

The given expression is-

$(x-10)(2x+6)(x^2-36)(x^2+10)(10x+20)=0$

This expression shows five factors. Using the zero-product property we can find all five solutions.

$\begin{gathered} x-10=0\rightarrow x=10 \\ 2x+6=0\rightarrow2x=-6\rightarrow x=-3 \\ x^2-36=0\rightarrow x^2=36\rightarrow x=\pm6 \\ x^2+10=0\rightarrow x^2=-10\rightarrow x=\sqrt[]{-10} \\ 10x+20=0\rightarrow10x=-20\rightarrow x=-\frac{20}{10}=-2 \end{gathered}$

Therefore, there are 4 real solutions and one imaginary solution, respectively, they are 10, -3, 6 and -6, -2, and the square root of negative 10.