Answer:
[tex]\textsf{Factored form}: \quad P(x)=\dfrac{2}{5}x(x-4)^2(x+2)[/tex]
[tex]\textsf{Standard form}: \quad P(x)=\dfrac{2}{5}x^4-\dfrac{12}{5}x^3+\dfrac{64}{5}x[/tex]
Step-by-step explanation:
Given information:
- P(x) = polynomial of degree 4.
- Root of multiplicity 2 at x = 4.
- Root of multiplicity 1 at x = 0.
- Root of multiplicity 1 at x = -2.
- Passes through point (5, 14).
The multiplicity of a root refers to the number of times the associated factor appears in the factored form of the equation of a polynomial.
Therefore:
[tex]\boxed{P(x)=ax(x-4)^2(x+2)}[/tex]
where a is a constant to be found.
To find the value of a, substitute the given point (5, 14) into the function and solve for a:
[tex]\begin{aligned}P(5)&=14\\\implies 5a(5-4)^2(5+2)&=14\\5a(1)^2(7)&=14\\5a(1)(7)&=14\\35a&=14\\a&=\dfrac{14}{35}\\a&=\dfrac{2}{5}\end{aligned}[/tex]
Therefore, the formula for P(x) in factored form is:
[tex]\boxed{P(x)=\dfrac{2}{5}x(x-4)^2(x+2)}[/tex]
Expand the factored form to write the formula in standard form:
[tex]\implies P(x)=\dfrac{2}{5}x(x+2)(x-4)^2[/tex]
[tex]\implies P(x)=\left(\dfrac{2}{5}x^2+\dfrac{4}{5}x\right)(x^2-8x+16)[/tex]
[tex]\implies P(x)=\dfrac{2}{5}x^4-\dfrac{16}{5}x^3+\dfrac{32}{5}x^2+\dfrac{4}{5}x^3-\dfrac{32}{5}x^2+\dfrac{64}{5}x[/tex]
[tex]\implies P(x)=\dfrac{2}{5}x^4-\dfrac{12}{5}x^3+\dfrac{64}{5}x[/tex]
