Question

Find roots of x³-5x²-16x+80 if two roots are equal but opposite in sign

Answer 1

We have
${x}^{3} - 5 {x}^{2} - 16x + 80 = 0 \\ \alpha + \beta + \gamma = 5 \\ \alpha + ( - \alpha ) + \gamma = 5 \\ \gamma = 5$
Thus, one root of the given equation is 5. Now we know that
$\alpha \beta \gamma = - 80 \\ \alpha ( - \alpha )5 = - 80 \\ - { \alpha }^{2} = - 16 \\ { \alpha }^{2} = 16 \\ \alpha = 4 \\ \alpha = - 4$
Thus, the roots of the given equation are
-4, 4, 5.
^_^

Answer 2

Answer:

Step-by-step explanation:

x^3-5x^2-16x+80 = x^2(x-5)-16(x-5)

                            = (x^2-16)(x-5)

                            =(x+4)(x-4)(x-5)

therefore the zeroes are 4, -4, 5