Question

question 9
the polynomial of degree 4, p(x) has a root of multiplicity 2 at x = 3 and roots of multiplicity 1 at x = 0 and x = - 4. it goes through the point (5,72).
find a formula for p(x).
p(x)=
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Explanation:

Step1: Write the polynomial in factored form

If \(x = r\) is a root of a polynomial, then \((x - r)\) is a factor. Given roots \(x = 3\) (multiplicity 2), \(x=0\) (multiplicity 1) and \(x=- 4\) (multiplicity 1), the polynomial \(P(x)\) can be written as \(P(x)=a(x - 3)^{2}(x-0)(x + 4)=a x(x - 3)^{2}(x + 4)\).

Step2: Find the value of \(a\)

Since the polynomial goes through the point \((5,72)\), substitute \(x = 5\) and \(P(x)=72\) into the equation.
\[

$$\begin{align*} 72&=a\times5\times(5 - 3)^{2}\times(5 + 4)\\ 72&=a\times5\times4\times9\\ 72&=a\times180\\ a&=\frac{72}{180}=\frac{2}{5} \end{align*}$$

\]

Step3: Write the final formula for \(P(x)\)

Substitute \(a=\frac{2}{5}\) back into \(P(x)\):
\[P(x)=\frac{2}{5}x(x - 3)^{2}(x + 4)\]
Expand \((x - 3)^{2}=x^{2}-6x + 9\), then \(P(x)=\frac{2}{5}x(x^{2}-6x + 9)(x + 4)\)
\[

$$\begin{align*} &=\frac{2}{5}x(x^{3}+4x^{2}-6x^{2}-24x + 9x+36)\\ &=\frac{2}{5}x(x^{3}-2x^{2}-15x + 36)\\ &=\frac{2}{5}(x^{4}-2x^{3}-15x^{2}+36x)\\ &=\frac{2}{5}x^{4}-\frac{4}{5}x^{3}-6x^{2}+\frac{72}{5}x \end{align*}$$

\]

Answer:

\(P(x)=\frac{2}{5}x(x - 3)^{2}(x + 4)\) (or \(\frac{2}{5}x^{4}-\frac{4}{5}x^{3}-6x^{2}+\frac{72}{5}x\))