Exercise 50 Page 588

As we can see, the coefficient of $x$ is the sum of the roots and the constant term is their product. With this in mind, let's complete the last column.

Roots	Factors	Equation
$2,$ $5$	$(x-2),$ $(x-5)$	$(x-2)(x-5)=0$ $x^2-7x+10=0$
$1,$ $9$	$(x-1),$ $(x-9)$	$(x-1)(x-9)=0$ $x^2-10x+9=0$
$\text{-}1,$ $3$	$(x+1),$ $(x-3)$	$(x+1)(x-3)=0$ $x^2-2x-3=0$
$0,$ $6$	$x,$ $(x-6)$	$x(x-6)=0$ $x^2-6x=0$
$\frac{1}{2},$ $7$	$\left(x-\frac{1}{2}\right),$ $(x-7)$	$\left(x-\frac{1}{2}\right)(x-7)=0$ $x^2-\frac{15}{2}x+\frac{7}{2}=0$
$\text{-}\frac{2}{3},$ $4$	$\left(x+\frac{2}{3}\right),$ $(x-4)$	$\left(x+\frac{2}{3}\right)(x-4)=0$ $x^2-\frac{10}{3}x-\frac{8}{3}=0$

To get rid of the fractions in the last two equations, we can multiply both sides of the equations by the denominator of the fractions.

Roots	Factors	Equation
$2,$ $5$	$(x-2),$ $(x-5)$	$x^2-7x+10=0$
$1,$ $9$	$(x-1),$ $(x-9)$	$x^2-10x+9=0$
$\text{-}1,$ $3$	$(x+1),$ $(x-3)$	$x^2-2x-3=0$
$0,$ $6$	$x,$ $(x-6)$	$x^2-6x=0$
$\frac{1}{2},$ $7$	$\left(x-\frac{1}{2}\right),$ $(x-7)$	$2x^2-15x+7=0$
$\text{-}\frac{2}{3},$ $4$	$\left(x+\frac{2}{3}\right),$ $(x-4)$	$3x^2-10x-8=0$

We will now multiply these factors and equate it to zero. The roots of this equation are $1,$ $2,$ and $3.$ This equation is not a quadratic equation, since the highest degree of its monomials is $3.$