Guided Example #3 - Polynomial Equations

By the factor theorem, P(x)has factors (x+2),(x-1), and (x-4). And since another linear factor (x-c) would produce a forth zero of P(x), no other factors of degree 1 exist.. Hence
${\displaystyle P\left(x\right)}$has the form
${\displaystyle P\left(x\right)=a(x+2)(x-1)(x-4)}$.

Since ${\displaystyle P\left(2\right)=16}$, we see that
${\displaystyle 16=a(2+2)(2-1)(2-4)}$

Solve for a
${\displaystyle 16=a\left(4\right)\left(1\right)(-2)}$
${\displaystyle a=-2}$

Therefore,
Multiplying the factors yield the polynomial
${\displaystyle P\left(x\right)=-2x^3+6x^2+12x-16.}$

By the factor theorem, P(x)has factors (x+2),(x-1), and (x-4). And since another linear factor (x-c) would produce a forth zero of P(x), no other factors of degree 1 exist.. Hence
${\displaystyle P\left(x\right)}$has the form
${\displaystyle P\left(x\right)=a(x+2)(x-1)(x-4)}$.

Since ${\displaystyle P\left(2\right)=16}$, we see that
${\displaystyle 16=a(2+2)(2-1)(2-4)}$

Solve for a
${\displaystyle 16=a\left(4\right)\left(1\right)(-2)}$
${\displaystyle a=-2}$

Therefore,
Multiplying the factors yield the polynomial
${\displaystyle P\left(x\right)=-2x^3+6x^2+12x-16.}$