Question

Consider the equation f(x)=x^2-4x-21

one of the zeros of the function is x=-3

what is the other zeros?

Answer 1

x = 7

Explanation:

Method 1:

`f(x)=ax^2+bx+c\\\\if\ x_1\ and\ x_2\ are\ the\ zeros,\ then\\\\f(x)=a(x-x_1)(x-x_2)\\\\\text{We have}\\\\f(x)=x^2-4x-21\to a=1\ \text{and}\ x_1=-3.\\\\\text{Substitute:}\\\\x^2-4x-21=1(x-(-3))(x-x_2)\\\\x^2-4x-21=(x+3)(x-x_2)\\\\\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\x^2-4x-21=(x)(x)+(x)(-x_2)+(3)(x)+(3)(-x_2)\\\\x^2-4x-21=x^2-xx_2+3x-3x_2\\\\x^2+(-4x)+(-21)=x^2+(-x_2+3)x+(-3x_2)\\\\\text{therefore}`

`-x_2+3=-4\ and\ -3x_2=-21\\\\-x_2+3=-4\qquad\text{subtract 3 from both sides}\\-x_2=-7\qquad\text{change the signs}\\x_2=7\\\\-3x_2=-21\qquad\text{divide both sides by (-3)}\\x_2=7`

Method 2:

`f(x)=x^2-4x-21\\\\x^2-4x-21=0\\\\x^2-7x+3x-21=0\\\\x(x-7)+3(x-7)+0\\\\(x-7)(x+3)=0\iff x-7=0\ \vee\ x+3=0\\\\x-7=0\qquad\text{add 7 to both sides}\\x=7\\\\x+3=0\qquad\text{subtract 3 from both sides}\\x=-3`