Determining cubic equation

ThinningKat

Just need assurance if i'm correct or not.

"Find the cubic equation that has zeros at -8, -4, and 13 if f(-10)=138[FONT=LatoWeb, Helvetica Neue, Helvetica, Arial, sans-serif]"[/FONT]

this is what i did but i think i'm not right. Can someone help correct it?

Cervesa

$f(x) = a(x+8)(x+4)(x-13)$

$f(-10)=138 \implies 138=a(-2)(-6)(-23) \implies a= \dfrac{138}{12(-23)}=-\dfrac{1}{2}$

Therefore, $f(x)=-\dfrac{1}{2}(x+8)(x+4)(x-13)$

1 person

Idea

$$\displaystyle f(-10)=138$$

therefore it should be

$$\displaystyle 138=a(x+8)(x+4)(x-13)$$ where $x=-10$

1 person