The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=2 and roots of multiplicity 1 at x=0 and x=-2. It goes thgough the point (5,157.5)

BolkowN

BolkowN

Answered question

2021-09-05

The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=2and roots of multiplicity 1 at x=0 and x=2. It goes thgough the point (5,157.5)
Find a formula for P(x).

Answer & Explanation

Margot Mill

Margot Mill

Skilled2021-09-06Added 106 answers

Given that the polynomials of degree 4,
If the polynomials has a root of multiplicity 2 at \(x=2\) then \(\displaystyle{\left({x}-{2}\right)}^{{2}}\) is a factor of given polynomials.
Multiplicity of 1 at \(x=0\) then “x” is a factor of given polynomials.
Multiplicity of 1 at \(x=-2\) then \((x+2)\) is a factor of given polynomials.
So,
\(\displaystyle{P}{\left({x}\right)}=\lambda\cdot{\left({x}-{2}\right)}^{{2}}\cdot{x}\cdot{\left({x}+{2}\right)}\)
The given polynomial \(P(x)\) passes through the points \((5,157.5)\) then,
\(\displaystyle{157.5}=\lambda\cdot{\left({5}-{2}\right)}^{{2}}\cdot{5}\cdot{\left({5}+{2}\right)}\)
\(\displaystyle{157.5}=\lambda\cdot{9}\cdot{5}\cdot{7}\)
\(\displaystyle{157.5}={315}\lambda\)
\(\displaystyle\lambda={\frac{{{157.5}}}{{{315}}}}\)
\(\displaystyle\lambda={0.5}\)
Substitute \(\displaystyle\lambda={0.5}\) in equation then,
\(\displaystyle{P}{\left({x}\right)}={0.5}{\left({x}-{2}\right)}^{{2}}\cdot{x}\cdot{\left({x}+{2}\right)}\)

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