Question

The polynomial of degree 4, P(x) has a root of multiplicity 2 at x = 4 and roots of multiplicity 1 at = 0 and x = -2. It goes through the point (5, 14). Find a formula for P(x). P(r)​

Answer 1

`\textsf{Factored form}: \quad P(x)=(2)/(5)x(x-4)^2(x+2)`

`\textsf{Standard form}: \quad P(x)=(2)/(5)x^4-(12)/(5)x^3+(64)/(5)x`

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:

`\boxed{P(x)=ax(x-4)^2(x+2)}`

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:

`\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&=(14)/(35)\\a&=(2)/(5)\end{aligned}`

Therefore, the formula for P(x) in factored form is:

`\boxed{P(x)=(2)/(5)x(x-4)^2(x+2)}`

Expand the factored form to write the formula in standard form:

`\implies P(x)=(2)/(5)x(x+2)(x-4)^2`

`\implies P(x)=\left((2)/(5)x^2+(4)/(5)x\right)(x^2-8x+16)`

`\implies P(x)=(2)/(5)x^4-(16)/(5)x^3+(32)/(5)x^2+(4)/(5)x^3-(32)/(5)x^2+(64)/(5)x`

`\implies P(x)=(2)/(5)x^4-(12)/(5)x^3+(64)/(5)x`